Calculus II Home

## 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 $\lim_{n\to \infty} \frac{n(n+3)}{(n+1)^2}=\lim_{n\to \infty} \frac{1\left(1+\frac{3}{n}\right)}{\left(1+\frac{1}{n}\right)^2}=1\neq 0.$

Last edited