An expression of the form \(a_1+a_2+...+a_n+...=\sum_{n=1}^\infty a_n\) is called an infinite series or simply a series, where \(a_n\) is the nth term of a sequence.
The nth partial sum \(s_n\) of an infinite series \(\sum_{n=1}^\infty a_n\) is \(s_n = a_1+a_2+...+a_n\).
Definition: An infinite series \(\sum_{n=1}^\infty a_n\) is convergent if \(\lim_{n\to\infty}s_{n}=s\) for some real number \(s\). The series is divergent if the limit does not exist (i.e., if it is not a real number).
If the infinite series \(\sum_{n=1}^\infty a_n\) is convergent if \(\lim_{n\to\infty}s_{n}=s\), then \(s\) is called the sum of the series and can be written as \(s=a_1+a_2+...+a_n+...\). If the series diverges, then it has no sum.
Geometric Series
The geometric series, \(\sum_{n=1}^\infty ar^{n-1}=a+ar+ar^2+ar^3+...=\sum_{n=0}^\infty ar^n\), converges if \(|r|<1\) has the sum \(\frac a{1-r}\) where \(a\) is the first term of the series and \(r\) is called the common ratio. The geometric series diverges if \(\left|r\right|\geq1\).
Example
Determine if the series converges or diverges. If the series converges, find its sum.
- \(\sum_{n=1}^\infty5^{-n}2^{n+1}\)
- \(\sum_{n=0}^\infty\frac{(-2)^{2n}}{3^{n-1}}\)
Solution
Rewrite the series in the form: \(\sum_{n=1}^\infty ar^{n-1}\)\[\begin{aligned}&\sum_{n=1}^{\infty}5^{-n}2^{n+1}=\sum_{n=1}^{\infty}\frac{2^{n+1}}{5^{n}}=\sum_{n=1}^{\infty}\frac{2^{n+1}}{5^{n}}\frac{5}{5}\frac{2^{2}}{2^{2}}\\&=\sum_{n=1}^{\infty}\frac{2^{n-1}}{5^{n-1}}\frac{4}{5}=\sum_{n=1}^{\infty}\frac{4}{5}\biggl(\frac{2}{5}\biggr)^{n-1}\end{aligned}\]
Then, \(a=\frac45\) and \(r=\frac{2}{5}<1\). Thus, this geometric series converges, and its sum is \(\frac a{1-r}=\frac{\frac45}{1-\frac25}=\frac43\).
Rewrite the series in the form: \(\sum_{n=0}^{\infty}ar^n\)
\(\sum_{n=0}^{\infty}\frac{(-2)^{2n}}{3^{n-1}}=\sum_{n=0}^{\infty}3\frac{4^{n}}{3^{n}}=\sum_{n=0}^{\infty}3\biggl(\frac{4}{3}\biggr)^{n}\)
Then, \(r=\frac{4}{3}>1\). Thus, this geometric series diverges, and it has no sum.
Telescoping Series
Use partial fraction decomposition for rational expressions to calculate the partial sum; for instance, to find the partial sum of \(\sum_{n=1}^{\infty}\frac{1}{n(n+1)}\) , use \(\frac1{n(n+1)}=\frac1n-\frac1{n+1}\). Then, expand the series up to its nth term to find the partial sum.
Example
Prove that the following infinite series converges, and find its sum:
\[\sum_{n=1}^{\infty}\frac{1}{n(n+1)}=\frac{1}{1(2)}+\frac{1}{2(3)}+\frac{1}{3(4)}+...+\frac{1}{n(n+1)}+...\]
Solution
The partial sum of this infinite series is
\[S_{n}=\sum_{i=1}^{n}\frac{1}{i(i+1)}=\frac{1}{1(2)}+\frac{1}{2(3)}+\frac{1}{3(4)}+...+\frac{1}{n(n+1)}\]
Using the partial fraction decomposition on the rational expression of the partial sum: \(\frac{1}{i(i+1)}=\frac{1}{i}-\frac{1}{i+1}\)
Then, once expanding the sum, the partial sum reduces to \(n\)
\[\begin{aligned}&S_{n}=\sum_{i=1}^{n}\left(\frac{1}{i}-\frac{1}{i+1}\right)=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...\\&+\left(\frac{1}{n-2}-\frac{1}{n-1}\right)+\left(\frac{1}{n-1}-\frac{1}{n}\right)+\left(\frac{1}{n}-\frac{1}{n+1}\right)\\&=1-\frac{1}{n+1}\end{aligned}\]
Finally, take the limit of the partial sum as \(n\) approaches to \(\infty \).
\[\lim\limits_{n\to\infty}S_n=\lim\limits_{n\to\infty}\biggl(1-\frac{1}{n+1}\biggr)=\lim\limits_{n\to\infty}\biggl(1-\frac{1/n}{1+1/n}\biggr)=1\].
p-series
Fact: The p-series \(\sum_{n=1}^\infty\frac1{n^p}\) is convergent if \(p>1\) and divergent if \(p\leq1\).
For example, \(\sum_{n=1}^\infty\frac1n\)(harmonic series) is divergent since it is a p-series with \(p=1\).