# Series

## Motivation: decimal approximation of transcendental numbers

• If you enter $$\pi$$ to your calculator, it's going to give you something like $\pi=3.1415926535\dotsc$
• We know that this is just an approximation, and the list of decimals could go on forever (you can get in TV if you can remember like a thousand of them.)
• We understand the equality as $\pi=3+\frac{1}{10}+\frac{4}{10^2}+\frac{1}{10^3}+\frac{5}{10^4}+\frac{9}{10^5}+\dotsb$
• There's an infinite sum on the right!
• Definition. Let $$\{a_n\}$$ be a sequence. Then the corresponding (infinite) series is the expression $a_1+a_2+a_3+\dotsb+a_n+\dotsb$
• Alternative shorthand notations are $$\sum_{n=1}^\infty a_n$$ or $$\sum a_n$$.

## Convergence for series

• Just as we say that $$3.14=3+\frac{1}{10}+\frac{4}{100}$$ approximates $$\pi$$, for an abritrary series $a_1+a_2+a_3+\dotsb+a_n+\dotsb$ and $$n\ge1$$, we'll call $s_n=a_1+a_2+a_3+\dotsb+a_n$ the $$n$$-th partial sum of the series, and we'll see if the sequence $$\{s_n\}$$ converges.
• Example.
• Consider the series $1+2+3+\dotsb+n+\dotsb$
• The partial sums are $s_n=1+2+3+\dotsb+n=\frac{n(n+1)}{2}.$
• We have $$\lim_{n\to\infty}\frac{n(n+1)}{2}=\infty$$, therefore the sequence $$\{s_n\}$$ diverges.

## Convergence for series

• Example.
• Consider the series $\frac12+\frac14+\frac18+\dotsb+\frac{1}{2^n}+\dotsb$
• Here, the partial sums are $s_n=\frac12+\frac14+\frac18+\dotsb+\frac{1}{2^n}=\frac{2^{n-1}+2^{n-2}+\dotsb+1}{2^n}=\frac{2^n-1}{(2-1)2^n}=1-2^{-n}.$
• Therefore, $$\lim_{n\to\infty}(1-2^{-n})=1$$ shows that the sequence $$\{s_n\}$$ has $$1$$ as its limit.
• Definition. Let $$a_1+a_2+a_3+\dotsb+a_n+\dotsb$$ be a series.
• If the sequence of partial sums $$\{s_n\}$$ satisfies $$\lim_{n\to\infty}s_n=s$$, then the series $$\sum a_n$$ is called convergent, and we write $a_1+a_2+a_3+\dotsb+a_n+\dotsb=s\text{ or }\sum_{n=1}^\infty a_n=s.$
• If the sequence of partial sums $$\{s_n\}$$ is divergent, then we'll say that the series $$\sum a_n$$ is divergent.

## Geometric series

• Let's now consider the geometric series: $a+ar+ar^2+\dotsb+ar^{n-1}+\dotsb$
• Each term is obtained from the preceding one by multiplying it by the common ratio $$r$$.
• If $$r=1$$, then we have $s_n=a+a+\dotsb+a=na.$ Since $$\lim_{n\to\infty}s_n=\infty$$, this series is divergent.
• If $$r\ne1$$, then we have $s_n=a(1+r+\dotsb+r^{n-1})=a\frac{r^n-1}{r-1}.$ If $$|r|>1$$, then the series is divergent. If $$|r|<1$$, then the series is convergent, and we have $$\sum_{n=0}^\infty ar^n=\frac{a}{1-r}$$.

## Geometric series

• Example. Find the sum of the geometric series $7-\frac{35}{8}+\frac{175}{64}-\dotsb$
• We can see that the first term is $$a=7$$, and then using the second term, we can find the common ratio $r=\frac{-\frac{35}{8}}{7}=-\frac58.$
• From this, we get $7-\frac{35}{8}+\frac{175}{64}-\dotsb=\frac{a}{1-r}=\frac{7}{\frac{13}{8}}=\frac{56}{13}.$
• Exercises. 11.2: 17, 19

## Examples

• Consider the number $0.1123123123123123\dotsc=0.1\overline{123}$
• We can write it as a geometric series plus an extra starting term: $\frac{1}{10}+\frac{123}{10^4}+\frac{123}{10^7}+\dotsb.$
• That is, we have $0.1\overline{123}=\frac{1}{10}+\frac{\frac{123}{10^4}}{1-10^{-3}}=\frac{1}{10}+\frac{123}{10^4-10}=\frac{999+123}{9990}=\frac{1122}{9990}.$
• Exercises. 11.2: 51, 53, 55
• Consider the series $x^0-x^1+x^2-\dotsb+(-1)^nx^n+\dotsb.$
• This is a geometric series with starting term $$x^0=1$$ and common ratio $$-x$$.
• Therefore, if $$|x|<1$$ then we have $\sum_{n=0}^\infty=\frac{1}{1+x},$
• and if $$|x|\ge1$$, then the series is divergent.
• Exercises. 11.2: 57, 59, 61

## Two canonical examples.

• Consider the series $$\sum_{n=1}^\infty\frac{1}{n(n+1)}$$.
• Via writing $$\frac{1}{i(i+1)}=\frac{1}{i}-\frac{1}{i+1}$$, the partials sums become telescoping sums: $s_n=\sum_{i=1}^n\left(\frac{1}{i}-\frac{1}{i+1}\right)=1-\frac12+\frac12-\frac13+\dotsb+\frac{1}{n}-\frac{1}{n+1}=1-\frac{1}{n+1}.$
• Therefore, we have $$\sum_{n=1}^\infty\frac{1}{n(n+1)}=1$$.
• Consider the harmonic series $$\sum_{n=1}^\infty\frac{1}{n}$$.
• We'll show that the subsequence $$s_2,s_4,s_8,\dotsc,s_{2^n},\dotsc$$ is divergent. This will imply that the original sequence is divergent too.
• We have: \begin{align*} s_2=&1+\frac12\\ s_4=&1+\frac12+\left(\frac13+\frac14\right)>1+\frac12+\left(\frac14+\frac14\right)=1+\frac22\\ s_8=&1+\frac12+\left(\frac13+\frac14\right)+\left(\frac15+\frac16+\frac17+\frac18\right)>1+\frac12+\left(\frac14+\frac14\right)+\left(\frac18+\frac18+\frac18+\frac18\right)=1+\frac32\\ \end{align*}
• Similarly, we get $$s_{2^n}>1+\frac{n}{2}$$. Since the sequence $$\left\{1+\frac{n}{2}\right\}$$ is divergent, so is $$s_{2^n}$$, and thus so is $$s_n$$.
• Exercises. 11.2: 43, 47.

## The divergence test, and operations on series.

• Theorem. Let $$\sum a_n$$ be a series. Suppose that it is convergent. Then we have $$\lim_{n\to\infty}a_n=0$$.
• Divergence test. Let $$\sum a_n$$ be a series. If it is not true that $$\lim_{n\to\infty}a_n=0$$, that is it either converges to a nonzero number or it is divergent, then the series $$\sum a_n$$ is divergent.
• Theorem. Let $$\sum a_n$$ and $$\sum b_n$$ be convergent series, and $$c$$ a constant. Then the series $$\sum(a_n+b_n)$$ and $$\sum ca_n$$ are also convergent, and we have \begin{align*} \sum(a_n+b_n)=&\sum a_n+\sum b_n,\\ \sum ca_n=&c\sum a_n. \end{align*}
• Warning. Just as with sequences, remember that these rules work only if the series $$\sum a_n$$ and $$\sum b_n$$ are convergent.
• Exercises. 11.2: 27, 28, 29, 33, 35, 37
• More exercises. 11.2: 69, 75, 81, 87, 89