Lecture 28:
Infinite Series
Definition. If \(X=(x_{n})\) is a sequence in \(\R\), then the infinite series (or simply series) generated by \(X\) is the sequence \(S = (s_{n})\) defined by \[s_{n} = \sum_{k=1}^{n}x_{k}.\]
- The numbers \(x_{n}\) are called the terms of the series.
- The number \(s_{k}\) is called the \(k\)th partial sum of the series.
- If the limit \(\lim S\) exists, then we say that the series is convergent, and this limit is called the sum of the series.
- If this limit does not exist, then we say that the series \(S\) is divergent.
Caution! A series is a type of sequence. These words are not interchangeable.
Infinite Series
We use the following symbols to denote both the series generated by \(X=(x_{n})\) and the sum of the series:
\[\sum(x_{n})\quad\text{or}\quad\sum x_{n}\quad\text{or}\quad\sum_{n=1}^{\infty}x_{n}\]
We often want to talk about sequences that are indexed \(\{k,k+1,k+2,k+3,\ldots\}\) and we write \((x_{n})_{n=k}^{\infty}\). For example
\[\left(\frac{1}{n^{2}-1}\right)_{n=2}^{\infty} = \left(\frac{1}{3},\frac{1}{8},\frac{1}{15},\ldots\right)\]
Similarly, we can have a series indexed by a different set, for example the series
\[\sum_{k=0}^{\infty}\frac{k+1}{k^{2}+1}\]
is the sequence
\[\left(1,1+1,1+1+\frac{3}{5},1+1+\frac{3}{5}+\frac{4}{10},1+1+\frac{3}{5}+\frac{4}{10}+\frac{5}{17},\ldots\right)\]
Example. Consider the series \[\sum_{n=1}^{\infty}\frac{1}{2^{n}}\]
Subtracting, we see that
\[s_{k} = \left(1+\frac{1}{2}+\frac{1}{2^2}+\cdots+\frac{1}{2^{k-1}}\right) - \left( \frac{1}{2}+\frac{1}{2^2}+\cdots + \frac{1}{2^{k}}\right)\]
\[ = 1-\frac{1}{2^{k}}\]
This is the series generated by \((\frac{1}{2^{n}}),\) and hence it is the sequence \(S=(s_{k})\) where \(s_{k}\) is the \(k\)th partial sum, that is \[s_{k} = \sum_{n=1}^{k}\frac{1}{2^{n}} = \frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots + \frac{1}{2^{k}}.\] The first few partial sums are
\[(s_{k}) = \left(\frac{1}{2},\frac{3}{4},\frac{7}{8},\frac{15}{16},\frac{31}{32},\ldots\right)\]
Multiplying \(s_{k}\) by \(2\) we see
\[2s_{k} = 1+\frac{1}{2}+\frac{1}{2^2}+\cdots+\frac{1}{2^{k-1}}.\]
\(=\lim(s_{k}) = 1\)
Example. Consider the series
\[\sum_{n=2}^{\infty}\frac{1}{5^{n}}\]
Subtracting, we see that
\[4s_{k} = \left(\frac{1}{5}+\frac{1}{5^2}+\cdots+\frac{1}{5^{k-1}}\right) - \left( \frac{1}{5^2}+\frac{1}{5^3}+\cdots + \frac{1}{5^{k}}\right)\]
\[ = \frac{1}{5}-\frac{1}{5^{k}}\]
Hence, the \(k\)th partial sum \(s_{k} = \dfrac{1}{4}\left(\dfrac{1}{5}-\dfrac{1}{5^{k}}\right).\)
.
This is the series generated by \((\frac{1}{5^{n}})_{n=2}^{\infty},\) and hence it is the sequence \(S=(s_{k})_{k=2}^{\infty}\) where \(s_{k}\) is the \(k\)th partial sum, that is
\[s_{k} = \sum_{n=2}^{k}\frac{1}{5^{n}} = \frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}+\cdots + \frac{1}{5^{k}}.\]
Multiplying by \(5\) we see
\[5s_{k} = \frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{k-1}}.\]
\[=\frac{1}{20}\]
Example. Consider the series
\[\sum_{n=2}^{\infty}\frac{1}{5^{n}}\]
Subtracting, we see that
\[4s_{k} = \left(\frac{1}{5}+\frac{1}{5^2}+\cdots+\frac{1}{5^{k-1}}\right) - \left( \frac{1}{5^2}+\frac{1}{5^3}+\cdots + \frac{1}{5^{k}}\right)\]
\[ = \frac{1}{5}-\frac{1}{5^{k}}\]
Hence, the \(k\)th partial sum \(s_{k} = \dfrac{1}{4}\left(\dfrac{1}{5}-\dfrac{1}{5^{k}}\right).\)
This is the series generated by \((\frac{1}{5^{n}})_{n=2}^{\infty},\) and hence it is the sequence \(S=(s_{k})_{k=2}^{\infty}\) where \(s_{k}\) is the \(k\)th partial sum, that is
\[s_{k} = \sum_{n=2}^{k}\frac{1}{5^{n}} = \frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}+\cdots + \frac{1}{5^{k}}.\]
Multiplying by \(5\) we see
\[5s_{k} = \frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{k-1}}.\]
Example. Consider the series
\[\sum_{n=1}^{\infty}\frac{1}{n(n+1)}\]
This is the series generated by \((\frac{1}{n(n+1)})_{n=1}^{\infty},\) and hence it is the sequence \(S=(s_{k})_{k=1}^{\infty}\) where \(s_{k}\) is the \(k\)th partial sum, that is
\[s_{k} = \sum_{n=1}^{k}\frac{1}{n(n+1)} = \frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\cdots + \frac{1}{k(k+1)}.\]
Note that for any \(n\in\N\) we have \[\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}\]
Hence, \[s_{k} = \sum_{n=1}^{k}\frac{1}{n(n+1)} = \sum_{n=1}^{k}\left(\frac{1}{n} - \frac{1}{n+1}\right) \]
\[=\left(1-\frac{1}{2}\right) + \left(\frac{1}{2}-\frac{1}{3}\right) + \left(\frac{1}{3}-\frac{1}{4}\right) + \cdots + \left(\frac{1}{k}-\frac{1}{k+1}\right)\]
\[=1-\frac{1}{k+1}\]
\[=1\]
Theorem 3.7.3 (the \(n\)th Term Test). If the series \(\sum x_{n}\) converges, then \(\lim(x_{n})=0\).
Proof. Suppose \(\sum x_{n}\) converges. For each \(k\in\N\) set \[s_{k} = \sum_{n=1}^{k} x_{n} = x_{1}+x_{2}+x_{3}+\cdots+x_{k}.\]
Let \(\varepsilon>0\) be given. By the Cauchy Convergence Criterion there is a natural number \(K\in\N\) such that \[|s_{k} - s_{m}|<\varepsilon\quad\text{for }k,m\geq K.\] If \(k\geq K+1\), then we have
\[|x_{k} - 0| = |s_{k} - s_{k-1}|<\varepsilon.\]
\(\Box\)
Note: \(\sum\frac{1}{n}\) does not converge, but \(\lim(\frac{1}{n})=0\). So the converse of this theorem is not true.
Lecture 28
By John Jasper
Lecture 28
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