Calculus – Sequences and series – Alternating series

An alternating series is a series whose terms are alternately positive and negative.

Examples:

1. \(\quad\displaystyle1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\),


2. \(\quad\displaystyle-\frac{1}{2}+\frac{2}{3}-\frac{3}{4}+\frac{4}{5}-\frac{5}{6}+\cdots=\sum_{n=1}^{\infty}(-1)^n\frac{n}{n+1}\).

Theorem: If the alternating series \(\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1}b_n\) with \(b_n > 0\) satisfies

1. \(b_{n+1}\leq b_n\) for all \(n\),


2. \(\lim\limits_{n\to\infty}b_n=0\),

then the series converges.

Proof:
We first consider the even partial sums \(s_2=b_1-b_2\geq0\), \(s_4=s_2+(b_3-b_4)\geq s_2\) and so on. In general:

\[s_{2n+2}=s_{2n}+(b_{2n+1}-b_{2n+2})\geq s_{2n}\quad\text{for}\quad n=1,2,3,\ldots.\]

Hence:

\[0\leq s_2\leq s_4\leq s_6\leq \cdots.\]

However, we also have:

\[s_{2n}=b_1-(b_2-b_3)-(b_4-b_5)-\cdots-(b_{2n-2}-b_{2n-1})-b_{2n}\leq b_1.\]

Hence the sequence \(\{s_{2n}\}\) of even partial sums is increasing and bounded above and therefore convergent. Say: \(\lim\limits_{n\to\infty}s_{2n}=s\). Then we have

\[\lim\limits_{n\to\infty}s_{2n+1}=\lim\limits_{n\to\infty}(s_{2n}+b_{2n+1})=\lim\limits_{n\to\infty}s_{2n}+\lim\limits_{n\to\infty}b_{2n+1}=s+0=s.\]

Since both the sequence of even and the sequence of odd partial sums converge to \(s\), we conclude that \(\lim\limits_{n\to\infty}s_n=s\) and this implies that the series converges.

Examples:

1) The series \(\displaystyle1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\) is convergent, since: \(\displaystyle\frac{1}{n+1}\leq\frac{1}{n}\) for all \(n\) and \(\lim\limits_{n\to\infty}\displaystyle\frac{1}{n}=0\).

2) The series \(\displaystyle\sum_{n=1}^{\infty}(-1)^n\frac{3n}{4n+1}\) is divergent, since: \(\displaystyle\lim\limits_{n\to\infty}\frac{3n}{4n+1}=\lim\limits_{n\to\infty}\frac{3}{4+\frac{1}{n}}=\frac{3}{4}\neq0\). This implies that the limit of the general term does not exist (and therefore is not equal to zero).

3) The series \(\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1}\frac{n^2}{n^3+1}\) is convergent, since: \(b_n=\displaystyle\frac{n^2}{n^3+1}=f(n)\) with \(f(x)=\displaystyle\frac{x^2}{x^3+1}\) and therefore

\[f'(x)=\frac{2x(x^3+1)-3x^4}{(x^3+1)^2}=\frac{x(2-x^3)}{(x^3+1)^2}<0\quad\text{for}\quad x>1.\]

This implies that \(b_{n+1}\leq b_n\) for \(n\geq2\). Further we have: \(\displaystyle\lim\limits_{n\to\infty}b_n=\lim\limits_{n\to\infty}\frac{n^2}{n^3+1} =\lim\limits_{n\to\infty}\frac{\frac{1}{n}}{1+\frac{1}{n^3}}=0\).

Estimating the sum of a series

Theorem: Let \(\displaystyle\sum(-1)^{n-1}b_n\) be an alternating series with \(b_n>0\), \(b_{n+1}\leq b_n\) for all \(n\) and \(\displaystyle\lim\limits_{n\to\infty}b_n=0\). If \(s_n\) denotes the \(n^{\text{th}}\) partial sum of the series and \(s=\lim\limits_{n\to\infty}s_n\), then we have for the remainder \(R_n\): \(|R_n|=|s-s_n|\leq b_{n+1}\).

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Last modified on March 15, 2021