p-Series

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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.