Calculus: Problems and Solutions http://thecalculusblog.com Hip calculus discussion, with problems and solutions Sun, 22 Apr 2012 21:02:05 +0000 en hourly 1 The Limit Comparison Test http://thecalculusblog.com/2011/07/04/the-limit-comparison-test/?utm_source=rss&utm_medium=rss&utm_campaign=the-limit-comparison-test http://thecalculusblog.com/2011/07/04/the-limit-comparison-test/#comments Mon, 04 Jul 2011 18:59:23 +0000 admin http://thecalculusblog.com/?p=273 Continue reading]]>

The comparison test lets us compare an unknown series to a simpler, known series. But it requires us to come up with inequalities that can be difficult to figure out. Worse, sometimes it just doesn’t quite work. For example, suppose we’re looking at the series

    \[ \sum_{n=2}^{\infty}\frac{1}{n^2-1}. \]

This series looks almost like a p-series with p=2, so we expect it to converge. However, if we try to use the comparison test to show that, we end up comparing n^2-1 with n^2. We have

    \[ n^2-1 < n^2, \]

so

    \[ \frac{1}{n^2-1} > \frac{1}{n^2}. \]

But that’s not the inequality we want! Using this inequality we can only show that our series has terms that are larger than the corresponding terms of a converging series. That doesn’t say anything about whether or not our series converges. So how can we proceed?

Statement of the Test

There are a couple of ways to salvage our argument, but one of the simplest is to use the Limit Comparison Test. This test says that if \sum a_n and \sum b_n are series with positive terms and the limit

    \[ \lim_{n\rightarrow\infty}\frac{a_n}{b_n} \]

exists and is nonzero, then either both \sum a_n and \sum b_n converge, or else both diverge.

In our example above, we want to compare the series

    \[ \sum_{n=2}^{\infty}\frac{1}{n^2-1} \]

and

    \[ \sum_{n=2}^{\infty}\frac{1}{n^2}. \]

These series both have positive terms, so we only need to calculate the limit

    \begin{align*} \lim_{n\rightarrow\infty}\frac{\frac{1}{n^2-1}}{\frac{1}{n^2}} & = \lim_{n\rightarrow\infty}\frac{n^2}{n^2-1}\\ & = 1. \end{align*}

By the limit comparison, both series above either converge or diverge. Since we already know that the second one converges (because it’s a p-series with p>1), the first one must converge, too.

Series Whose Terms Are Quotients of Polynomials

The limit comparison test works great with series whose terms are quotients of polynomials. As an example, let’s take a look at the series

    \[ \sum_{i=1}^{\infty}\frac{4i^2+i+1}{i^3+i^2+2i+1}. \]

What are the largest terms in the numerator and denominator? In the numerator, 4i^2 is growing the fastest. In the denominator it’s i^3. So we expect the series to behave like the simpler series

    \[ \sum_{i=1}^{\infty}\frac{4i^2}{i^3} = 4\sum_{i=1}^{\infty}\frac{1}{n}. \]

This series is the harmonic series, so it diverges.

Both series have positive terms, so to use the limit comparison test we need to compute the limit

    \begin{align*} \lim_{i\rightarrow\infty}\frac{\frac{4i^2+i+1}{i^3+i^2+2i+1}}{\frac{4}{i}} & = \lim_{i\rightarrow\infty}\frac{4i^3+i^2+i}{4i^3+4i^2+8i+4} \\ & = 1. \end{align*}

Since the limit exists and isn’t zero, the limit comparison test tells us that both series must diverge.

]]> http://thecalculusblog.com/2011/07/04/the-limit-comparison-test/feed/ 0 The Comparison Test For Series http://thecalculusblog.com/2011/07/01/the-comparison-test-for-series/?utm_source=rss&utm_medium=rss&utm_campaign=the-comparison-test-for-series http://thecalculusblog.com/2011/07/01/the-comparison-test-for-series/#comments Sat, 02 Jul 2011 00:38:55 +0000 admin http://thecalculusblog.com/?p=268 Continue reading]]>

Sometimes we’re given a series that is very close to something we can deal with. For example, the series

    \[ \sum_{n=1}^{\infty}\frac{1}{n^2+1} \]

is very close to the p-series

    \[ \sum_{n=1}^{\infty}\frac{1}{n^2}. \]

In cases like this, we can use the comparison test to decide whether the series converges. First, remember that the second series above converges since it’s a p-series with p=2. Second, notice that

    \[ n^2+1 > n^2, \]

so

    \[ \frac{1}{n^2+1}<\frac{1}{n^2} \]

(if you don’t understand or don’t believe it, plug in some numbers for n.)

Since each term of the first series is smaller than each term of the second series, we expect that the first series converges since the second one does. That’s exactly what the comparison test tells us.

Details of The Comparison Test

Suppose that \sum a_n and \sum b_n are series with positive terms. If a_n\leq b_n and \sum b_n converges, so does \sum a_n. On the other hand, if a_n\leq b_n and \sum a_n diverges, then so does \sum b_n.

Also, notice that not every of the first series needs to be less than the corresponding series. It just needs to be true eventually. For example, the series

    \[ 1+ \frac{1}{2}+ \frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\dots\]

converges, and we can use this to show that

    \[ 3+2+1+\frac{1}{9}+\frac{1}{17} + \frac{1}{33} +\dots\]

does, too, even though the first few terms are actually bigger than the corresponding terms of the first series.

The comparison test is useful especially in certain situations. Take a look at these problems using the comparison test:

Problem: Show that

    \[ \sum_{n=1}^{\infty}\frac{|\sin n|}{n^2} \]

converges.

Solution: Remember that \sin x is never bigger than 1 and never less than -1. That means that

    \[ 0\leq |\sin n|\leq 1\]

for every integer n. Dividing each of these terms by n^2, we get the inequality

    \[ 0\leq\frac{|\sin n|}{n^2}\leq \frac{1}{n^2}. \]

We know that \sum_{n=1}^{\infty}\frac{1}{n^2} converges because it’s a p-series with p=2. Therefore the given series converges by the comparison test.

In the above problem, there was a \sin n that was making things complicated. We used the comparison test to get rid of it and simplify the series we were looking at. That’s the general philosophy you should use when you’re thinking about the comparison test. Always compare to something simpler than what you’re looking at. Here’s another problem where we can use the comparison test to simply the series we’re thinking about:

Problem: Show that

    \[ \sum_{j=1}^{\infty}\frac{e^{-j}}{j^4} \]

converges.
Solution:The function e^{-x} is positive and decreasing, and when x\geq 0 its value is less than or equal to 1. So, we have

    \[ 0< e^{-j}\leq 1 \]

when j\geq 1. Dividing this inequality by j^4, we get

    \[ 0< \frac{e^{-j}}{j^4}\leq \frac{1}{j^4}. \]

But the series \sum_{j=1}^{\infty}\frac{1}{j^4} converges. Therefore the original series does, too, by the comparison test.

]]> http://thecalculusblog.com/2011/07/01/the-comparison-test-for-series/feed/ 2 The Harmonic Series http://thecalculusblog.com/2011/05/24/the-harmonic-series/?utm_source=rss&utm_medium=rss&utm_campaign=the-harmonic-series http://thecalculusblog.com/2011/05/24/the-harmonic-series/#comments Wed, 25 May 2011 03:49:20 +0000 admin http://thecalculusblog.com/?p=265 Continue reading]]>

A special type of p-series is the harmonic series. This is a p-series with p=1, in other words the series

    \[ \sum_{n=1}^{\infty}\frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \dots\]

Since p-series only converge when p>1, the harmonic series diverges.

Notice that the harmonic series is an example of a series whose terms tend to zero (\lim_{n\rightarrow\infty}\frac{1}{n} = 0) but the series doesn’t converge. Remember this if you’re ever tempted to use the divergence test to conclude that a series converges.

Also note that related series like

    \[ \sum_{n=1}^{\infty} \frac{3}{n} \]

and

    \[ \sum_{n=1}^{\infty} \frac{-1}{n} \]

also diverge. You can show this using the integral test, for example.

]]> http://thecalculusblog.com/2011/05/24/the-harmonic-series/feed/ 0 p-Series http://thecalculusblog.com/2011/05/23/p-series/?utm_source=rss&utm_medium=rss&utm_campaign=p-series http://thecalculusblog.com/2011/05/23/p-series/#comments Mon, 23 May 2011 20:47:34 +0000 admin http://thecalculusblog.com/?p=260 Continue reading]]>

Using the integral test, we can look at an important class of infinite series: the so-called p-series. These are series of the form

    \[ \sum_{n=1}^{\infty}\frac{1}{n^p} \]

for some number p. Here are some examples:

    \[ \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \dots \]

(a p-series with p=2),

    \[ \sum_{i=1}^{\infty} \frac{1}{\sqrt{i}} \]

(a p-series with p=1/2), and

    \[ 1 + 2 + 3 + 4 + 5 + 6 + \dots \]

(a p-series with p=-1).

Here’s the important thing to remember about p-series: A p-series converges if and only if p> 1. That means the first example above converges, and the other two don’t.

How can we show that a p-series converges if and only if p>1? Using the integral test, of course. Let f(x) = \frac{1}{x^p}. Then it’s easy to check that f is continuous on [1,\infty) and f is decreasing on that interval. Therefore the series

    \[ \sum_{n=1}^{\infty} \frac{1}{n^p} \]

converges if and only if the integral

    \[ \int_1^{\infty}\frac{1}{x^p}\, dx\]

does. Let’s evaluate the integral, assuming that p\neq 0:

    \begin{align*} \int_1^{\infty}\frac{1}{x^p}\, dx &= \lim_{t\rightarrow\infty}\int_1^t\frac{1}{x^p}\,dx\\ &= \lim_{t\rightarrow\infty} \left. \frac{1}{-p+1}x^{-p+1}\right|_1^t\\ & = \lim_{t\rightarrow\infty} \frac{1}{-p+1}t^{-p+1} - \frac{1}{-p+1}. \end{align*}

How does t^{-p+1} behave as t\rightarrow\infty? If the exponent is positive, it blows up (approaches infinity). If the exponent is negative, however, it approaches zero and the integral converges. Therefore the series converges if -p+1<0, or in other words, if p>1.

What happens if p=0? You can use the integral test in this case, too. I’ll leave it up to you to show that the integral diverges, and therefore so does the sum.

]]> http://thecalculusblog.com/2011/05/23/p-series/feed/ 0 Absolute Maxima And Minima Problems http://thecalculusblog.com/2011/05/20/absolute-maxima-and-minima-problems/?utm_source=rss&utm_medium=rss&utm_campaign=absolute-maxima-and-minima-problems http://thecalculusblog.com/2011/05/20/absolute-maxima-and-minima-problems/#comments Fri, 20 May 2011 21:29:57 +0000 admin http://thecalculusblog.com/?p=247 Continue reading]]>

Finding the absolute maxima and minima of functions (the biggest and smallest values a function attains) is important for many real world calculus applications. For example, companies want to maximize their profits, engineers want to minimize the drag on an airplane, computer scientists want to get your youtube video from the server to your computer as fast as possible. This type of mathematics can also address questions like why soap bubbles take the shapes they do, and these types of problems very quickly lead to the frontiers of mathematical physics. But enough of that, let’s look at some examples of finding the absolute max and min of functions.

Have a look at the function below, which is 3x^4-16x^3+18x^2 on the interval [-1,4].

Rendered by QuickLaTeX.com

Let’s consider the minima of this function first. It looks like this function has a local minimum at x=0 and an absolute minimum at x=3. Absolute here means that this is the smallest value the function attains on that interval. Local means that if you zoom way in it’s a minimum, but clearly the function gets smaller at x=3.

How could we do this if we didn’t have a graph of the function? If all we’re given is f(x)=3x^4-16x^3+18x^2 on the interval [-1,4], we need to find the critical points:

    \begin{align*} f'(x) &= 12x^3-3\cdot 16x^3+36x\\ & = 12x(x^2-4x+3)\\ & = 12x(x-1)(x-3) \end{align*}

Therefore f'(x) = 0 at x=0, x=1, and x=3. The first and last of these are the minima we saw above. It looks from the graph like x=1 is a maximum. Is that the absolute maximum of the function? No, clearly not! The function is bigger at x=-1 and x=4.

This is an important point: to find the absolute maximum and minimum of a function on a closed interval [a,b], first find the critical points of the function. Plug in these values to the function, and also plug in the endpoints x=a and x=b. The biggest number is the absolute maximum and the smallest number is the absolute minimum of the function.

It turns out that continuous functions on a closed, bounded interval always have an absolute maximum and an absolute minimum. However, that’s not true if the interval isn’t closed or isn’t bounded. For example, e^x has no maximum value on [0,\infty). It just keeps growing and growing. It’s also not hard to think of an example of a function that isn’t continuous which has no maximum or minimum value on an interval.

Here are some more problems to think about:

Q: What are the absolute maximum and minimum values of e^{-x} on the interval [-1, 1]? What about on the interval [0,\infty)?

Answer: if f(x)=e^{-x}, then f'(x) = -e^{-x}. This function is never zero, so it has no critical points. Since f is decreasing, on the interval [-1, 1] the absolute maximum is e and the absolute minimum is e^{-1}.

Q: Find the absolute maximum and minimum values of f(x)=1-x^2 when 0<x\leq 2. What about when -1\leq x\leq 2?

Answer: f'(x) =-2x here, so f has a critical point at 0. This isn’t in the set of x such that 0<x\leq 2, though. Since f'<0 when 0<x\leq 2, f is decreasing there. Since f has its maximum at 0, but this point isn’t in the domain, f has no absolute maximum on the domain. It does have an absolute minimum at f(2) = -3.

On the other hand, on the domain -1\leq x\leq 2, we need to check the critical point x=0 as well as the endpoints x=-1 and x=2. Computing, we get f(-1) = 0, f(0) = 1, and f(2) = -3. That means the absolute maximum of f is 1 and the absolute minimum is -3.

]]> http://thecalculusblog.com/2011/05/20/absolute-maxima-and-minima-problems/feed/ 1 The Integral Test For Series http://thecalculusblog.com/2011/05/13/integral-test-for-series/?utm_source=rss&utm_medium=rss&utm_campaign=integral-test-for-series http://thecalculusblog.com/2011/05/13/integral-test-for-series/#comments Fri, 13 May 2011 13:31:48 +0000 admin http://thecalculusblog.com/?p=241 Continue reading]]>

One way we can determine whether or not a series converges is to view it as an area and see if the area is finite. Of course, we know how to find the area under a function (by integrating), so this suggests that integration might somehow be useful. That’s what’s involved in the integral test for series.

Suppose we want to know whether the series

    \[ \sum_{n=1}^{\infty} a_n \]

converges or not. Define a function f(x) by replacing all the n’s in a_n with x’s. If f is continuous, positive, and decreasing, and if

    \[ \int_1^{\infty} f(x)\,dx \]

converges, so does the sum. Conversely, if the integral diverges, the sum does, too.

The integral test is very powerful when you can integrate the terms of the series you’re looking at. Let’s have an example: does the series

    \[ \sum_{n=1}^{\infty}\frac{1}{n^2} \]

converge?

Answer: we can apply the integral test. Define f(x)=\frac{1}{x^2}. Then a_n = f(n), f is continuous, positive, and decreasing on (1, \infty), and so the integral test applies. We have

    \begin{align*} \int_1^{\infty}\frac{1}{x^2}\,dx & = \lim_{t\rightarrow\infty}\int_1^t x^{-2}\,dx \\ & = \lim_{t\rightarrow\infty} \left.-x^{-1}\right|_1^t \\ & = \lim_{t\rightarrow\infty} -\frac{1}{t} + \frac{1}{1}\\ & = 1. \end{align*}

The integral converges, and therefore so does the series.

Another common example: does

    \[ \sum_{k=2}^{\infty}\frac{1}{k\ln k} \]

converge?

Answer: define f(x) = \frac{1}{x\ln x}. Then f is continuous and positive on (2,\infty). Notice that x\ln x is increasing on this interval, and so its reciprocal is decreasing. Therefore the integral test applies:

    \[ \int_2^{\infty} \frac{1}{x\ln x}\,dx & = \lim_{t\rightarrow\infty}\int_2^t \frac{1}{x\ln x}\, dx, \]

and now we can let u=\ln x in the integral, so that du = 1/x\,dx. With the new limits, the integral is

    \begin{align*} \lim_{t\rightarrow\infty} \int_{\ln 2}^{\ln t}\frac{1}{u}\,du & = \lim_{t\rightarrow\infty}\left.\ln |u|\right|_{\ln 2}^{\ln t} \\ & = \lim_{t\rightarrow\infty}\ln(\ln t) - \ln |\ln 2|\\ & = \infty. \end{align*}

Since the integral diverges, so does the given series.

]]> http://thecalculusblog.com/2011/05/13/integral-test-for-series/feed/ 0 Partial Fractions (Or, Partial Fraction Decomposition) http://thecalculusblog.com/2011/05/11/partial-fractions-or-partial-fraction-decomposition/?utm_source=rss&utm_medium=rss&utm_campaign=partial-fractions-or-partial-fraction-decomposition http://thecalculusblog.com/2011/05/11/partial-fractions-or-partial-fraction-decomposition/#comments Wed, 11 May 2011 05:19:20 +0000 admin http://thecalculusblog.com/?p=230 Continue reading]]>

There are some situations when it’s useful to write a rational function (a quotient of polynomials) as a sum of simpler such functions. As an example, we can see by cross multiplying that

    \begin{align*} \frac{1}{x-1}+\frac{1}{x+1} & = \frac{(x+1) + (x-1)}{(x-1)(x+1)}\\ & = \frac{2x}{x^2-1}. \end{align*}

Partial fraction decomposition (often called just partial fractions) allows us to go backward, to figure out how to write 2n/(n^2-1) as a sum. It works like this: given a rational function whose denominator has higher degree than its numerator, we factor the denominator. Then we write the fraction as a sum of fractions whose denominators are the factors of the original denominator. The numerators of these fractions are yet to be determined, so we usually write them with variables A, B, and so on, whose values we have to figure out. Then we add together these fractions, getting an equation in the variables A, B, and so on by setting the numerator equal to the original numerator. Finally, we use this equation to solve for the A, B, and so on.

It sounds complicated, but it’s usually pretty simple. Let’s take a look at an example:

    \[ \frac{1}{x^2-5x+6} \]

We can factor x^2-5x+6 = (x-2)(x-3). So we want to find numbers A and B such that

    \[ \frac{1}{x^2-5x+6} = \frac{A}{x-2} + \frac{B}{x-3}. \]

To do this, cross multiply to get

    \[ \frac{A}{x-2} + \frac{B}{x-3} = \frac{A(x-3) + B(x-2)}{(x-2)(x-3)}. \]

Since we want this to be equal to the original fraction, and their denominators are equal, their numerators must be equal as well. Therefore we have the equation A(x-3) + B(x-2) = 1. This holds for every value of x. It’s convenient to put in the values x=3 and x=2. When x=3, we get B=1. When x=2, we get A=-1. Putting these back in for A and B we get the desired decomposition

    \[ \frac{1}{x^2-5x+6} = \frac{-1}{x-2} + \frac{1}{x-3}. \]

Not too hard, huh?

What if there’s a repeated factor in the denominator? For example, how can we write

    \[ \frac{x+1}{(x-1)^2} \]

as a sum of simpler fractions? When there are repeated factors in the denominator, we have to repeat these factors in the sum as well: here we’ll take

    \[ \frac{x+1}{(x-1)^2} = \frac{A}{x-1} + \frac{B}{(x-1)^2} \]

as our decomposition. Cross multiplying we get

    \begin{align*} \frac{A}{x-1} + \frac{B}{(x-1)^2} &= \frac{A(x-1)^2 + B(x-1)}{(x-1)^3}\\ & = \frac{A(x-1) + B}{(x-1)^2}. \end{align*}

The numerator must be equal to the original numerator, so we see that A(x-1)+B = x+1. Putting in x=1 we get B=2, and so A=1. Therefore

    \[ \frac{x+1}{(x-1)^2} = \frac{1}{x-1} + \frac{2}{(x-1)^2}. \]

Finally, we need to see what happens if there is an irreducible polynomial (one with no real roots, such as x^2 + 1) in the denominator. Here we have to take the numerator to be a degree 1 function instead of a constant. For example, let’s take a look at

    \[ \frac{2x+2}{(x-1)(x^2+1)}. \]

We want to write this as

    \[ \frac{A}{x-1}+\frac{Bx+C}{x^2+1}. \]

Note the linear term in the numerator of the second fraction. Cross multiplying we get

    \[ \frac{A}{x-1}+\frac{Bx+C}{x^2+1} = \frac{A(x^2+1) + (Bx+C)(x-1)}{(x-1)(x^2+1)}. \]

The numerator of this is the same as the numerator of the original fraction, so (A+B)x^2 + (C-B)x + A-C = 2x+2. Solving for A, B, and C, we get A=2, B=-2, and C=0, so that

    \[ \frac{2x+2}{(x-1)(x^2+1)} = \frac{2}{x-1}+\frac{-2}{x^2+1}. \]

You can read more about partial fractions at wikipedia.

]]> http://thecalculusblog.com/2011/05/11/partial-fractions-or-partial-fraction-decomposition/feed/ 0 Telescoping Series http://thecalculusblog.com/2011/05/11/telescoping-series/?utm_source=rss&utm_medium=rss&utm_campaign=telescoping-series http://thecalculusblog.com/2011/05/11/telescoping-series/#comments Wed, 11 May 2011 05:15:01 +0000 admin http://thecalculusblog.com/?p=228 Continue reading]]>

Some series look complicated, but they’re really simple. Consider this one:

    \[ 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \frac{1}{4} -\dots \]

It’s pretty clear that every term except the 1 cancels, so the series should converge to 1. (In fact, it’s not too hard to show this. Can you write an expression for the partial sum s_n?)

Telescoping series are similar to the one above. Take a look at the series

    \[ \sum_{n=1}^{\infty}\frac{1}{n^2+n}. \]

Using partial fractions we can rewrite the terms of the series as

    \[ \frac{1}{n^2+n} = \frac{1}{n} - \frac{1}{n+1}, \]

so the new series becomes

    \[ \sum_{n=1}^{\infty}\frac{1}{n}-\frac{1}{n+1}. \]

Let’s take a look at the first few terms of this series:

    \[ \sum_{n=1}^{\infty}\frac{1}{n}-\frac{1}{n+1} = \left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+\dots. \]

Check out what’s happening: all the terms cancel except for the first one (1). You might want to write down a couple more terms to convince yourself of this.

Any time you can use partial fractions to simplify the terms of a series, you should see if the series is telescoping. If it is, then it converges, and you can see what value it converges to by looking at which terms don’t cancel.

Let’s look at another example: by using partial fractions on the series

    \[ \sum_{i=1}^{\infty}\frac{6}{i^2+2i}, \]

we see that the sum is equal to

    \[ \sum_{i=1}^{\infty}\frac{3}{n}-\frac{3}{n+2}. \]

The first few terms of this series are

    \[ \left(\frac{3}{1}-\frac{3}{3}\right)+\left(\frac{3}{2}-\frac{3}{4}\right)+\left(\frac{3}{3}-\frac{3}{5}\right)+\left(\frac{3}{4}-\frac{3}{6}\right)+\dots, \]

and we can see that all terms cancel except for \frac{3}{1}+\frac{3}{2}. (You should take a minute to convince yourself that all the other terms really do cance.) Therefore this series converges to this value.

]]> http://thecalculusblog.com/2011/05/11/telescoping-series/feed/ 0 Geometric Series http://thecalculusblog.com/2011/05/09/geometric-series/?utm_source=rss&utm_medium=rss&utm_campaign=geometric-series http://thecalculusblog.com/2011/05/09/geometric-series/#comments Tue, 10 May 2011 04:01:14 +0000 admin http://thecalculusblog.com/?p=225 Continue reading]]>

The first type of series we usually learn about is called a geometric series. They’re so named because the ratio or successive terms is constant, but who cares about that. What do they look like? Here are two examples:

    \[ 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots \]

and

    \[ 3 + \frac{3}{5} + \frac{3}{25} + \frac{3}{125} + \dots\]

are both geometric series. In sum notation, we can write these series as

    \[ \sum_{n=0}^{\infty}\left(\frac{1}{2}\right)^n \]

and

    \[ \sum_{n=0}^{\infty}3\left(\frac{1}{5}\right)^n \]

respectively.

Another example of a geometric series is

    \[ \sum_{i=0}^{\infty} 7\cdot 2^i = 3 + 7\cdot 2 + 7\cdot 2^2 + 7\cdot 2^3 + \dots \]

However, we know by the divergence test that this series doesn’t converge, so it’s not as interesting. (Does the divergence test really work on this series? Check it!)

In general, a geometric series is one whose terms are a constant times a constant raised to the power of the variable you’re summing over. They look like this:

    \[ \sum_{n=0}^{\infty} a r^n \]

where a and r are constants (numbers). Note that we could also write this series as

    \[ \sum_{n=1}^{\infty} ar^{n-1} \]

Geometric series are special. We can determine whether or not they converge, but we can also say exactly what their sum is. Usually this is very, very difficult. But not so with geometric series. We can use this observation to find the partial sums: since a_n = ar^{n-1} for a geometric series, s_n = a + ar + ar^2 + \dots + ar^{n-1}. Multiplying by r and subtracting, we get

    \begin{align*} s_n - rs_n & = a + ar + ar^2 + \dots + ar^{n-1} \\ &\quad - (ar + ar^2 + \dots + ar^{n-1} + ar^n)\\ & = a - ar^n. \end{align*}

Solving for s_n, we obtain

    \[ s_n = \frac{a-ar^n}{1-r}. \]

What happens to this sequence as n\rightarrow\infty? If |r|<1, r^n just gets smaller and smaller, approaching zero. So, when |r|<1, s_n\rightarrow \frac{a}{1-r}. On the other hand, when |r|\geq 1, the test for divergence shows that the series diverges. In summary, we have

    \[ \sum_{n=0}^{\infty} ar^n = \frac{a}{1-r} \]

if |r|<1, and the series diverges otherwise.

Sometimes geometric series are hiding and we have to do some algebra to get them to show themselves. Consider this example:

    \[ \sum_{n=0}^{\infty}\frac{3^{n+1}}{2^{2n}}. \]

Doesn’t really look like a geometric series, quite. But we can rearrange the terms like so:

    \begin{align*} \frac{3^{n+1}}{2^{2n}} & = \frac{3\cdot 3^n}{(2^2)^n}\\ & = 3\frac{3^n}{4^n}\\ & = 3\left(\frac{3}{4}\right)^n. \end{align*}

So

    \[ \sum_{n=0}^{\infty}\frac{3^{n+1}}{2^{2n}} = \sum_{n=0}^{\infty} 3\left(\frac{3}{4}\right)^n, \]

and now we can tell that this is a geometric series. Furthermore, since 0<\frac{3}{4}<1, it converges to the value

    \[ \frac{3}{1-\frac{3}{4}} = 12. \]

Here are some more problems involving geometric series:

  1. The series

        \[ \sum_{n=1}^{\infty}\frac{1}{n^2} \]

    is not geometric, because n (the variable being summed over) doesn’t appear in the exponent of a constant. Similarly

        \[ \sum_{n=1}^{\infty}\frac{3}{n2^n} \]

    isn’t geometric because there’s an extra n (the one that isn’t in an exponent).

  2. The series

        \[ \sum_{i=1}^{\infty} \frac{1}{2^{i-1}} \]

    is geometric with a=1 and r=1/2. It converges because |r|<1, and its value is 2.

  3. The series

        \[ \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \dots \]

    is geometric. Here a=1/3, r=1/3, the series converges, and its value is 1/2.

  4. The series

        \[ \sum_{k=0}^{\infty} \frac{5^k}{4^{k+2}} \]

    is geometric. What are a and r? Does the series converge? If so, to what value?

]]> http://thecalculusblog.com/2011/05/09/geometric-series/feed/ 0 The Divergence Test http://thecalculusblog.com/2011/05/06/the-divergence-test/?utm_source=rss&utm_medium=rss&utm_campaign=the-divergence-test http://thecalculusblog.com/2011/05/06/the-divergence-test/#comments Sat, 07 May 2011 03:29:27 +0000 admin http://thecalculusblog.com/?p=217 Continue reading]]>

Now that we can find the sum of an infinite series when it converges, we’d like to be able to tell if a given series really does converge. Often it’s not easy to tell. But there’s one great test that can sometimes tell us when a series doesn’t converge. It’s called

The Divergence Test


Let

    \[ \sum_{n=1}^{\infty} a_n \]

be a series. Then if \lim_{n\rightarrow\infty} a_n\neq 0 or doesn’t exist, then the series diverges.

Note: the divergence test can only tell you that a series diverges. It can never tell you that a given sequence converges. That’s why it’s called the “divergence test.” This is super duper important. Don’t make the mistake of writing something like “this sequence converges because \lim_{n\rightarrow\infty} a_n = 0” on your test. This is one of the most common mistakes students make with series.

Let’s look at a couple of examples. Consider the sequence

    \[ \sum_{n=1}^{\infty} \left(1-\frac{1}{n}\right). \]

Let’s do the divergence test. We must compute: \lim_{n\rightarrow\infty}\left(1-\frac{1}{n}\right) = 1\neq 0, so by the divergence test the series diverges. Why does this make sense? Consider large values of n, so that 1/n is very small. Then the series looks almost like this

    \[ \ldots + 1 + 1 + 1 + 1 + \dots \]

I say “almost” because those numbers a just a little smaller than 1. However, it’s pretty clear that such a series won’t converge: the partial sums are getting larger and larger.

Problem: Consider the series

    \[ \sum_{n=1}^{\infty} \frac{1}{n}. \]

What does the divergence test say about this series?

Answer: Nothing! We can compute the limit: \lim_{n\rightarrow\infty}\frac{1}{n} = 0, but this doesn’t tell us that the series converges. It also doesn’t tell us that the series diverges. It doesn’t tell us anything at all! Mystery!

]]> http://thecalculusblog.com/2011/05/06/the-divergence-test/feed/ 0