How To Find The Sum Of An Infinite Series

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Click to see Calculus Sequences and Series: Problems and Solutions

Now that we know what a series is we need to figure out how to assign a sum to them. The trick is this: we turn the series into a sequence, and if that sequence converges we’ll call its limit the sum of the series.

In detail, given a series

    \[ \sum_{i=1}^{\infty} a_i\]

define a sequence called the sequence of partial sums:

    \begin{align*} s_1&=a_1\\ s_2&=a_1+a_2\\ s_3&=a_1+a_2+a_3\\ s_4&=a_1+a_2+a_3+a_4\\ \vdots& \end{align*}

Can you see why the sequence s_1, s_2, s_3,\dots is called the sequence of “partial sums?” This sequence is how we assign a value to the above series. If s = \lim_{n\rightarrow\infty} s_n exists, we say that the series converges to s. Otherwise, if the limit doesn’t exist or equals \infty or -\infty, we say that the series diverges.

Notice that s_2-s_1 = a_2, s_3-s_2=a_3, s_4-s_3=a_4, and in general s_n-s{n-1}=a_n. That means all the information about the series is contained in the sequence of partial sums. If somebody tells you, for example, that

    \[ s_n = \frac{1}{n^2}, \]

you can compute

    \begin{align*} a_n & = s_n-s_{n-1}\\ & = \frac{1}{n^2}-\frac{1}{(n-1)^2} \end{align*}

to get the terms of the series.

A few more problems are shown below. For even more problems with complete solutions, check out my book Calculus Sequences and Series: Problems and Solutions available at Amazon.

Problem 1: Suppose that the sequence of partial sums of a certain series is

    \[ s_i = 3 - \frac{n}{n^2+1}. \]

Does the series converge? If so, to what value?

Answer: The series converges if lim_{i\rightarrow\infty} s_i does. But this limit is just 3. So the answer is that the series converges to 3.

Problem 2: Does the series

    \[ \sum_{i=1}^{\infty} (-1)^i \]

converge?

Answer 2:Compute the sequence of partial sums: s_1=-1, s_2=-1+1=0, s_3=-1+1-1=-1, and so on. So the sequence of partial sums is -1, 0, -1, 0, -1, 0,\dots. This sequence clearly doesn’t converge, and therefore neither does the series in the problem.