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Series

     

A series is the sum of the terms of a sequence \(\{a_n\}\) which is denoted by \(\displaystyle\sum_{n=1}^{\infty} a_n\).

Example.

  1. \(\displaystyle\sum_{n=1}^{\infty} \frac{1}{2^n}=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots\) (a geometric series).

  2. \(\displaystyle\sum_{n=1}^{\infty} \frac{1}{n(n+1)}=\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\cdots\) (a telescoping series).


A series \(\displaystyle\sum_{n=1}^{\infty} a_n\) as an area

Definition. For the series \(\displaystyle\sum_{n=1}^{\infty} a_n\), the \(n\)th partial sum, denoted by \(s_n\), is \[s_n=\sum_{i=1}^n a_i=a_1+a_2+a_3+\cdots+a_n.\] The series \(\displaystyle\sum_{n=1}^{\infty} a_n\) converges to a real number \(s\) if the sequence \(\{ s_n \}\) of partial sums of \(\displaystyle\sum_{n=1}^{\infty} a_n\) converges to \(s\). We write \[\sum_{n=1}^{\infty} a_n=\lim_{n\to \infty} s_n=\lim_{n\to \infty}\sum_{i=1}^n a_i.\] A series is called convergent if it converges. Otherwise it is divergent.


Example. Consider the telescoping series \(\displaystyle\sum_{n=1}^{\infty} \frac{1}{n(n+1)}=\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\cdots\).
Note that \(s_1=\displaystyle\frac{1}{1\cdot 2},\,s_2=\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3},\, s_3=\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}, \ldots.\) \[\begin{align*} s_n &= \sum_{i=1}^n \frac{1}{i(i+1)}\\ &= \sum_{i=1}^n \left( \frac{1}{i}-\frac{1}{i+1} \right)\\ &= \left( \frac{1}{1}-\frac{1}{2} \right)+\left( \frac{1}{2}-\frac{1}{3} \right)+\cdots+\left( \frac{1}{n-1}-\frac{1}{n} \right)+\left( \frac{1}{n}-\frac{1}{n+1} \right)\\ &= 1-\frac{1}{n+1}. \end{align*}\] \[\sum_{n=1}^{\infty} \frac{1}{n(n+1)}=\lim_{n\to \infty} s_n=\lim_{n\to \infty} \left(1-\frac{1}{n+1} \right)=1.\]


Theorem. The geometric series \(\displaystyle\sum_{n=1}^{\infty} ar^{n-1}=a+ar+ar^2+\cdots\) is convergent if \(|r|<1\), i.e., \(-1< r < 1\) and divergent if \(|r|\geq 1\). In particular, \[ \sum_{n=1}^{\infty} ar^{n-1}= \frac{a}{1-r}, \;\; |r|<1. \]

\[ s_n=a+ar+ar^2+\cdots+ar^{n-1} \implies rs_n=ar+ar^2+ar^3+\cdots+ar^n. \] Then \[ s_n-rs_n=a-ar^n \implies (1-r)s_n=a(1-r^n) \implies s_n=a\frac{1-r^n}{1-r} \] Since \(|r|< 1\), \[\sum_{n=1}^{\infty} ar^{n-1}=\lim_{n\to \infty} s_n=\lim_{n\to \infty} a\frac{1-r^n}{1-r}=\frac{a}{1-r}.\]

Example.

  1. \(\displaystyle\sum_{n=1}^{\infty} \frac{1}{2^n}=\sum_{n=1}^{\infty} \frac{1}{2}\left(\frac{1}{2}\right)^{n-1}= \frac{\frac{1}{2}}{1-\frac{1}{2}}=1\) where the first term \(a=\frac{1}{2}\) and the common ratio \(r=\frac{1}{2}\) is in \((-1,1)\).

  2. \(\displaystyle\sum_{n=1}^{\infty} \frac{e^n}{3^{n-1}}=\sum_{n=1}^{\infty} e\left(\frac{e}{3}\right)^{n-1}= \frac{e}{1-\frac{e}{3}}=\frac{3e}{3-e}\) where the first term \(a=e\) and the common ratio \(r=\frac{e}{3}\) is in \((-1,1)\).

  3. We show that \(0.\bar{3}=\frac{1}{3}\).
    \[0.\bar{3}=0.333\cdots=0.3+0.03+0.003+\cdots=\displaystyle\sum_{n=1}^{\infty} \frac{3}{10}\left(\frac{1}{10}\right)^{n-1}= \frac{\frac{3}{10}}{1-\frac{1}{10}}=\frac{1}{3},\] where \(a=\frac{3}{10}\) and \(r=\frac{1}{10}\) is in \((-1,1)\).

  4. \(\displaystyle\sum_{n=0}^{\infty} x^n=\frac{1}{1-x}\) whenever \(|x|<1\). Note that \(\displaystyle\sum_{n=0}^{\infty} x^n=\sum_{n=1}^{\infty} x^{n-1}\) where \(a=1\) and \(r=x\) is in \((-1,1)\).

Algebra for series:
Theorem. If \(\displaystyle\sum_{n=1}^{\infty} a_n\) and \(\displaystyle\sum_{n=1}^{\infty} b_n\) are convergent, then so are the following series: \[ \sum_{n=1}^{\infty} (a_n+ b_n),\; \sum_{n=1}^{\infty} (a_n-b_n),\;\sum_{n=1}^{\infty} ca_n, \] for all real numbers \(c\). In particular, \[ \sum_{n=1}^{\infty} (a_n+ b_n)=\sum_{n=1}^{\infty} a_n + \sum_{n=1}^{\infty} b_n,\; \sum_{n=1}^{\infty} (a_n-b_n)=\sum_{n=1}^{\infty} a_n-\sum_{n=1}^{\infty} b_n,\;\sum_{n=1}^{\infty} ca_n=c\sum_{n=1}^{\infty} a_n. \]


Example. Find \(\displaystyle\sum_{n=1}^{\infty} \left( \frac{2}{n(n+1)}+\frac{1}{e^n} \right)\).

Solution. We know that \(\displaystyle\sum_{n=1}^{\infty} \frac{1}{n(n+1)}=1\) (the telescoping series) and \(\displaystyle\sum_{n=1}^{\infty} \frac{1}{e^n}=\frac{\frac{1}{e}}{1-\frac{1}{e}}=\frac{1}{e-1}\) (the geometric series with \(a=r=\frac{1}{e}\)). Then \[\sum_{n=1}^{\infty} \left( \frac{2}{n(n+1)}+\frac{1}{e^n} \right)=2\sum_{n=1}^{\infty} \frac{1}{n(n+1)}+\sum_{n=1}^{\infty}\frac{1}{e^n}=2\cdot 1+\frac{1}{e-1}=\frac{2e-1}{e-1}. \]


Theorem. If \(\displaystyle\sum_{n=1}^{\infty} a_n\) is convergent, then \(\displaystyle\lim_{n\to \infty} a_n=0\).

Suppose \(\displaystyle\sum_{n=1}^{\infty} a_n=\lim_{n\to \infty} s_n=s\). Since \(a_n=s_n-s_{n-1}\), \[ \lim_{n\to \infty} a_n=\lim_{n\to \infty} (s_n-s_{n-1})=\lim_{n\to \infty} s_n-\lim_{n\to \infty} s_{n-1}=s-s=0. \]

Note that the converse of the preceding theorem is not true.
Example. The harmonic series \(\displaystyle\sum_{n=1}^{\infty} \frac{1}{n}\) is divergent but \(\displaystyle\lim_{n\to \infty} \frac{1}{n}=0\).

\[\begin{align*} s_{2^n}&=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{2^n}\\ &=1+\left(\frac{1}{2}\right)+ \left(\frac{1}{3}+\frac{1}{4}\right) + \left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right) \cdots+ \left(\frac{1}{2^{n-1}+1}+\frac{1}{2^{n-1}+2}+\cdots+\frac{1}{2^n}\right)\\ &\geq 1+2^0\left(\frac{1}{2}\right)+ 2^1\left(\frac{1}{4}\right) + 2^2\left(\frac{1}{8}\right) \cdots+ 2^{n-1}\left(\frac{1}{2^n}\right)\\ &=1+\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)+\cdots+\left(\frac{1}{2}\right)\\ &=1+n\left(\frac{1}{2}\right) \end{align*}\] Thus \(s_{2^n}\geq 1+\frac{n}{2}\) for all natural numbers \(n\). Then \(\displaystyle\lim_{n\to \infty} s_{2^n}=\infty\) which implies \[\displaystyle\sum_{n=1}^{\infty} \frac{1}{n}=\lim_{n\to \infty} s_n=\infty.\]

By the contrapositive of the preceding theorem, we have the following result:
Theorem.(Divergence Test) If \(\displaystyle\lim_{n\to \infty} a_n \neq 0\) or the limit does not exist, the \(\displaystyle\sum_{n=1}^{\infty} a_n\) is divergent.

Example. \(\displaystyle\sum_{n=1}^{\infty} \frac{n(n+3)}{(n+1)^2}\) is divergent by the Divergence Test because \[\lim_{n\to \infty} \frac{n(n+3)}{(n+1)^2}=\lim_{n\to \infty} \frac{1\left(1+\frac{3}{n}\right)}{\left(1+\frac{1}{n}\right)^2}=1\neq 0. \]


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