Introduction

Welcome to this guide on Alternating Series Error Bound, this article will guide you through determining the error bound of a convergent alternating series

Alternating Series Error Bound

Consider an alternating series:

$\sum _{n=1}^{\infty }(-1{)}^{n+1}\frac{1}{n}$

This series converges, but how far is the first n-th (say 5) terms from the actual convergence value? To solve this problem, we introduce the Alternating Series Error Bound Theorem:

If an alternating series converges, then $\Sigma =\left|S-S_n\right|\le \left|a_{n+1}\right|$

Where $\Sigma$ is the error.

Intuitive Understanding

This section will introduce an intuitive understanding (not rigorous proof) to the theorem introduced earlier. Consider this series:

$\sum _{n=0}^{\infty }(-1{)}^n{a}_n=a_0-a_1+a_2-a_3+a_4-a_5+...=L$

Let the sum of the first 4 terms of this series be $S_n$, thus

$\sum _{n=0}^{\infty }(-1{)}^n{a}_n=S_n+a_4-a_5+...=L$

Since this series converge, $a_{n+1}<a_n$, which means that $\left|a_4\right|>\left|-a_5+a_6-...\right|$, or $\left|a_4\right|>\left|\Sigma -a_4\right|$thus error $\Sigma =\left|a_4-a_5+a_6-...\right|<2\left|a_4\right|$, meaning that

$S_n-a_{n+1}\le L\le S_n+a_{n+1}$

Example 1:

Estimate the convergence value of this series (give an estimation on upper and lower bound):

$\sum _{n=1}^{\infty }(-1{)}^{n+1}\frac{1}{n}$

Let's first add up the first 5 term of the series:

$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}=\frac{47}{60}$

By the introduced theorem, we have $\Sigma \le \left|a_6\right|$, or $\Sigma \le \left|\frac{1}{6}\right|$, meaning that:

$\frac{47}{60}-\frac{1}{6}\le \sum _{n=1}^{\infty }(-1{)}^{n+1}\frac{1}{n}\le \frac{47}{60}+\frac{1}{6}$

$\frac{37}{60}\le \sum _{n=1}^{\infty }(-1{)}^{n+1}\frac{1}{n}\le \frac{19}{20}$

This series actually converges to $\ln (2)\approx 0.693$, which is with in the error bound

Practice

Estimate the error bound of the following series using the first 5 terms