- 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 \(\sum a_n\).

- 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.

*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*.

- If the sequence of partial sums \(\{s_n\}\) satisfies \(\lim_{n\to\infty}s_n=s\), then the series \(\sum a_n\) is called

- 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 \(\sum_{n=0}^\infty ar^n=\frac{a}{1-r}\).

- Each term is obtained from the preceding one by multiplying it by the

- 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

- 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

- 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 \(\sum_{n=1}^\infty\frac{1}{n(n+1)}=1\).

- Via writing \(\frac{1}{i(i+1)}=\frac{1}{i}-\frac{1}{i+1}\), the partials sums become
- 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 \(s_n\).

- We'll show that the
- Exercises. 11.2: 43, 47.

*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