Series

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A series is summation of a sequence.

Let’s jump right in.

A series is a sum of an infinite sequence.

Let’s start our investigation on this topic with a question (a little unfair, I know!).

Can the sum of an infinite number of terms be a finite value?

no sometimes

As we will see, the answer is “sometimes.” Believe it or not, you have been working with infinite sums of numbers (also called series) for a long time. Consider the number $\frac {1}{3} = 0.3333333333\dots .$ This is the infinite sum of the geometric sequence $(a_n)_{n=1}^\infty$ where $a_n = \frac {3}{10^{n}}$ , as $\begin{align*} \sum_{n=1}^\infty 3\cdot \frac{1}{10^{n}} &= 0.3 + 0.03+0.003+ 0.0003+ 0.00003+ \cdots\\ &= \frac{3}{10} + \frac{3}{10^2} + \frac{3}{10^3} + \frac{3}{10^4} + \frac{3}{10^5} + \cdots\\ &=\frac{1}{3}. \end{align*}$ We can sum other geometric series to finite values as well. Consider $\sum _{n=1}^\infty \left (\frac {1}{4}\right )^n = \frac {1}{4} + \left (\frac {1}{4}\right )^2 + \left (\frac {1}{4}\right )^3 + \left (\frac {1}{4}\right )^4 + \cdots$ A very clever method of summing this sequence is as follows. Consider an equilateral triangle with area $1$ :

We can break this triangle into $4$ congruent triangles, each of area $1/4$ :

We can break the upper triangle into $4$ more congruent triangles, each with area $(1/4)^2$ :

Repeating this process, we find:

where the area of the shaded triangles is our geometric series: $\frac {1}{4} + \left (\frac {1}{4}\right )^2 + \left (\frac {1}{4}\right )^3 + \left (\frac {1}{4}\right )^4 + \cdots$ The area is clearly finite (it is between $0$ and $1$ !). What is the shaded area? Well, if you look at any “row” of the triangle, we’ve shaded in exactly one third of the row. Hence we’ve shaded in one third of the entire area, so we see $\frac {1}{3}=\frac {1}{4} + \left (\frac {1}{4}\right )^2 + \left (\frac {1}{4}\right )^3 + \left (\frac {1}{4}\right )^4 + \cdots$ While this is a very cool argument, it doesn’t generalize well. Let’s consider an argument that will apply to more settings.

Explain why $\frac {1}{3}=\frac {1}{4} + \left (\frac {1}{4}\right )^2 + \left (\frac {1}{4}\right )^3 + \left (\frac {1}{4}\right )^4 + \cdots$

Here is the idea: first, “name” your sum $S$ . $S = \frac {1}{4} + \left (\frac {1}{4}\right )^2 + \left (\frac {1}{4}\right )^3 + \left (\frac {1}{4}\right )^4 + \cdots$ Now, multiply $S$ by $\frac {1}{4}$ and write this suggestively under $S$ . $\begin{align*} S &= \frac{1}{4} + \left(\frac{1}{4}\right)^2 + \left(\frac{1}{4}\right)^3 + \left(\frac{1}{4}\right)^4 + \cdots\\ \left(\frac{1}{4}\right)S &= \left(\frac{1}{4}\right)^2 + \left(\frac{1}{4}\right)^3 + \left(\frac{1}{4}\right)^4 + \left(\frac{1}{4}\right)^5+ \cdots \end{align*}$ subtracting the lower line from the upper line we find $\begin{align*} S - \left(\frac{1}{4}\right)S &= \frac{1}{4}\\ S(1-\frac{1}{4}) &= \frac{1}{4}\\ S &= \frac{\frac{1}{4}}{1-\frac{1}{4}}\\ S &= \frac{1}{3}. \end{align*}$

This is a good method for understanding infinite sums of geometric sequences (assuming you know the sequence sums to a finite value).

To make this precise, we need some definitions.

Let $(a_k)$ be a sequence.

(a): The sum $\sum _{k=1}^\infty a_k$ is an infinite series (or, simply series).
(b): Let $S_n = \sum _{k=1}^n a_k$ ; the sequence $(S_n)$ is the sequence of $n$ th partial sums of $(a_k)$ .
(c): If $\lim _{n\to \infty } S_n = L$ , we say the series $\sum _{k=1}^\infty a_k$ converges to $L$ , and we write $\sum _{k=1}^\infty a_k = L$ .
(d): If $\lim _{n\to \infty } S_n$ diverges, the series $\sum _{k=1}^\infty a_k$ diverges.

Look back at the definition again. Previously, we’ve been using notation like $(a_n)$ to define our sequences. Suddenly, we’ve switched the index to a $k$ instead of an $n$ in some places. This notation is to help you keep in mind the differences between the two at work, here: $(a_k)$ and $(S_n)$ . The first sequence gives us the terms of our series; we are not usually concerned about its limit. We do want to consider the limit of the sequence of partial sums $(S_n)$ , since this will give us the sum of the series.

Using our new terminology, what is the behavior of the series we considered above? The series $\sum _{n=1}^\infty \left (\frac {1}{4}\right )^n$ convergesdiverges , and $\sum _{n=1}^\infty \left (\frac {1}{4}\right )^n = \answer {1/3}$ .

Geometric series

A geometric series is a series of the form $\sum _{k=0}^\infty ar^k$ for some real numbers $a \ne 0$ and $r$ .

We started this section with two different geometric series that sum to the same value. One reason geometric series are important is that they have nice convergence properties.

Consider the geometric series $\sum _{k=0}^\infty r^k$ . Compute the value of the $n$ th partial sum.

Let’s try to get a feel for what is going on by writing out the first few partial sums. $\begin{align*} S_1 &= r^0 \\ S_2 &= r^0 + r^1\\ S_3 &= r^0 + r^1 + r^2\\ &\vdots\\ S_n &= r^0 + r^1 + r^2 + \dots + r^{n-1} \end{align*}$ To compute $S_n$ , use the same trick we used before, multiply $S_n$ by $r$ $\begin{align*} S_n &= 1 + r^1 + r^2 + \dots + r^{n-1}\\ r S_n &= r^1 + r^2 + \dots + r^{n-1} + r^n \end{align*}$ and subtract to find $\begin{align*} S_n - r S_n &= 1 - r^n\\ S_n(1-r) &= 1 - r^n\\ S_n &= \frac{1 - r^n}{1-r}. \end{align*}$ Since $S_n$ is always a finite sum, there is no issue with manipulating it the way we did.

From our work above, we see that the $n$ th partial sum of the geometric series $a_n = r^n$ is $S_n = \sum _{k=0}^{n-1} r^k= \frac {1 - r^n}{1-r}.$ Using this fact, we can prove a theorem.

A geometric series $\sum _{k= 0}^\infty r^k$ converges if and only if $|r| < 1$ and when $|r|<1$ , $\sum _{k=0}^\infty r^k = \frac {1}{1-r}.$

Remember, to say that a series $\sum _{k= 0}^\infty r^k$ converges means that the limit of the partial sums converges. We know already know how to compute the $n$ th partial sum $S_n = \frac {1 - r^n}{1-r},$ so write with me: $\begin{align*} \sum_{k= 0}^\infty r^k &= \lim_{n\to\infty}S_n \\ &= \lim_{n\to\infty}\frac{1 - r^n}{1-r}. \end{align*}$ If $-1<r<1$ , then we have $\lim _{n\to \infty }\frac {1 - r^n}{1-r} = \frac {1}{1-r},$ and otherwise the sequence diverges.

According to the theorem above the series $\sum _{k=0}^\infty \left (\frac {1}{4}\right )^k = 1 + \frac {1}{4} + \left (\frac {1}{4}\right )^2 + \left (\frac {1}{4}\right )^3 + \cdots$ converges, and $\sum _{k=0}^\infty \left (\frac {1}{4}\right )^k = \frac {1}{1-1/4} = 4/3.$ This concurs with our introductory example; while there we got a sum of $1/3$ , we skipped the first term of $1$ .

You must pay close attention to how the series is indexed, since $\sum _{k=0}^n r^k \ne \sum _{k=1}^n r^k.$

Which of the following series converge?

$\sum _{k=0}^\infty \left (\frac {3}{2}\right )^k$ $\sum _{k=0}^\infty \left (\frac {-2}{3}\right )^k$ $\sum _{k=9}^\infty \left (\frac {1}{7}\right )^k$ $\sum _{k=1}^\infty (-1)^k$ $\sum _{k=-9}^\infty \left (\frac {1}{2}\right )^k$

The initial index doesn’t matter as far as convergence is concerned, it is the “tail” of the sequence that determines convergence.

Let’s use our new tools to find the sums of some geometric series.

If the series $\sum _{n=-2}^\infty \left (\frac {3}{4}\right )^n$ converges, find its sum.

Since the common ratio between the terms of this series is $\answer [given]{3/4}$ , we see that this series convergesdiverges . Write with me. $\begin{align*} S &= \left(\frac{3}{4}\right)^{-2} + \left(\frac{3}{4}\right)^{-1} + \left(\frac{3}{4}\right)^{0} + \cdots\\ \left(\frac{3}{4}\right) S &= \left(\frac{3}{4}\right)^{-1} + \left(\frac{3}{4}\right)^{0} + \left(\frac{3}{4}\right)^{1} + \cdots \end{align*}$ Subtracting these two lines we find $\begin{align*} S - \left(\frac{3}{4}\right) S &= \left(\frac{3}{4}\right)^{-2}\\ S\left(1-\frac{3}{4}\right) &= \left(\frac{3}{4}\right)^{-2}\\ S &= \frac{(3/4)^{-2}}{1-3/4}. \end{align*}$

If the series $\sum _{n=8}^\infty \left (\frac {-1}{2}\right )^n$ converges, find its sum.

Since the common ratio between the terms of this series is $\answer [given]{-1/2}$ , we see that this series convergesdiverges . Write with me. $\begin{align*} S &= \left(\frac{-1}{2}\right)^{8} + \left(\frac{-1}{2}\right)^{9} + \left(\frac{-1}{2}\right)^{10} + \cdots\\ \left(\frac{-1}{2}\right) S &= \left(\frac{-1}{2}\right)^{9} + \left(\frac{-1}{2}\right)^{10} + \left(\frac{-1}{2}\right)^{11} + \cdots \end{align*}$ Subtracting these two lines we find $\begin{align*} S - \left(\frac{-1}{2}\right) S &= \left(\frac{-1}{2}\right)^{8}\\ S\left(1-\frac{-1}{2}\right) &= \left(\frac{-1}{2}\right)^{8}\\ S &= \frac{(-1/2)^{8}}{1+1/2}. \end{align*}$

Connections to decimals

Remember how we pointed out that $\frac {1}{3} = 0.3333333333\dots$ is a geometric series? We can use our techniques for summing geometric series to find fractions equal to given decimals.

Find a fraction equal to $0.47474747474747\dots$

Do this exactly the same way as the examples we’ve done before. Write $\begin{align*} N &= 0.47474747474747\dots\\ 100 N &= 47.47474747474747\dots \end{align*}$ Now subtract the top line from the bottom line, to find $\begin{align*} 100N - N &= 47\\ N(100-1) &= 47\\ N &= \frac{47}{99} \end{align*}$

Find a fraction equal to $9.42764864864864864864\dots$

Do this exactly the same way as the examples we’ve done before. Write $\begin{align*} N &= 9.42764864864864864864\dots\\ 1000 N &= 9427.64864864864864864\dots \end{align*}$ Now subtract the top line from the bottom line, to find $\begin{align*} 1000N - N &= \\ N(1000-1) &= 9418.221\\ N &= \frac{9418.221}{999}. \end{align*}$ Multiplying the numerator and denominator of this fraction by $1000$ , our fraction will be $\frac {9418221}{999000}.$

Telescoping series

Telescoping series are infinite sums that “collapse” to more manageable expressions. Let’s see an example.

Evaluate the sum $\sum _{k=1}^\infty \left (\frac {1}{k}-\frac {1}{k+1}\right ).$

It will help to write down the first few partial sums of this series.

Note how most of the terms in each partial sum are canceled out! In general, we see that $S_n = 1-\frac {1}{n+1}$ . The sequence $(S_n)$ converges, as $\lim _{n\to \infty }S_n = \lim _{n\to \infty }\left (1-\frac 1{n+1}\right ) = 1$ , and so we conclude that $\sum _{n=1}^\infty \left (\frac 1n-\frac 1{n+1}\right ) = 1$ .

We’ve just seen an example of a telescoping series. Informally, a telescoping series is one in which the partial sums reduce to just a finite number of terms. The partial sum $S_n$ did not contain $n$ terms, but rather just two: $S_n = 1 - 1/(n+1).$ The example above gives us a moral: when possible, seek a way to write an explicit formula for the $n$ th partial sum $S_n$ . This makes evaluating the limit $\lim _{n\to \infty } S_n$ much more approachable. We do so in the next example.

Evaluate the sum: $\sum _{n=1}^\infty \frac {2}{n^2+2n}$

We can use partial fractions to write $\frac 2{n^2+2n} = \frac 1n-\frac 1{n+2}.$ Expressing the terms of $(S_n)$ is now more instructive.

We again have a telescoping series. In each partial sum, most of the terms cancel and we obtain the formula $S_n =1+\frac 12-\frac 1{n+1}-\frac 1{n+2}.$ Taking limits allows us to determine the convergence of the series: $\lim _{n\to \infty }S_n = \lim _{n\to \infty } \left (1+\frac 12-\frac 1{n+1}-\frac 1{n+2}\right ) = \frac 32,$ so $\sum _{n=1}^\infty \frac 1{n^2+2n} = \frac 32.$

Evaluate the sum: $\sum _{n=1}^\infty \ln \left (\frac {n+1}{n}\right )$

We begin by writing the first few partial sums of the series: $\begin{align*} S_1 &= \ln\left(2\right) \\ S_2 &= \ln\left(2\right)+\ln\left(\frac32\right) \\ S_3 &= \ln\left(2\right)+\ln\left(\frac32\right)+\ln\left(\frac43\right) \\ S_4 &= \ln\left(2\right)+\ln\left(\frac32\right)+\ln\left(\frac43\right)+\ln\left(\frac54\right) \end{align*}$ At first, this does not seem helpful, but recall the logarithmic rule $\ln (x)+\ln (y) = \ln (x\cdot y)$ . Applying this rule to $S_4$ gives: $\begin{align*} S_4 &= \ln\left(2\right)+\ln\left(\frac32\right)+\ln\left(\frac43\right)+\ln\left(\frac54\right) \\ &= \ln\left(\frac21\cdot\frac32\cdot\frac43\cdot\frac54\right)\\ &= \ln\left(5\right). \end{align*}$ We can conclude that $(S_n) =\ln (n+1).$ This sequence does not converge, as $\lim _{n\to \infty }S_n=\infty$ . Therefore $\sum _{n=1}^\infty \ln \left (\frac {n+1}{n}\right )=\infty$ ; the series diverges.

Properties of sums

We are learning about a new mathematical object, the series. As done before, we apply “old” mathematics to this new topic.

Properties of Infinite Series Let $\sum _{n=1}^\infty a_n = L,\quad \sum _{n=1}^\infty b_n =K,$ and let $c$ be a constant.

(a): Constant Multiple Rule: $\sum _{n=1}^\infty c\cdot a_n = c\cdot \sum _{n=1}^\infty a_n = c\cdot L.$
(b): Sum/Difference Rule: $\sum _{n=1}^\infty \big (a_n\pm b_n\big ) = \sum _{n=1}^\infty a_n \pm \sum _{n=1}^\infty b_n = L \pm K.$

Notice, of course, that we’re working with convergent series in this theorem. The results don’t necessarily hold if $\sum _{n=1}^\infty a_n$ or $\sum _{n=1}^\infty b_n$ are divergent!