Question

Test the following series for convergence.

3.$\sum _{n=1}^{\infty }\frac{{(-1)}^n}{n^2}$

Short answer

The series$\sum _{n=1}^{\infty }\frac{{(-1)}^n}{n^2}$ converges.

Step-by-step solution

Given information

It is given that the series is$\sum _{n=1}^{\infty }\frac{{(-1)}^n}{n^2}$

Convergence Test

According to the convergence test for alternating series: An alternating series converges if the absolute value of the terms decreases steadily to zero, that is, if $\left|a_{n+1}\right|\le \left|a_n\right|$, and$\underset{n\to \infty }{\lim }{a}_n=0$.

Apply the properties of convergence test

Let $a_n=\frac{{(-1)}^n}{n^2}$.

Then, $a_{n+1}=\frac{{(-1)}^{n+1}}{{\left(n+1\right)}^2}$.

Apply the convergence test as $\left|a_{n+1}\right|\le \left|a_n\right|$.

$\begin{array}{c}\left|\frac{{(-1)}^{n+1}}{{(n+1)}^2}\right|\le \left|\frac{{(-1)}^n}{{\left(n\right)}^2}\right|\\ \frac{1}{{(n+1)}^2}<\frac{1}{n^2}\end{array}$

Since the denominator is bigger, therefore the overall term is lesser. So, the first property is satisfied.

Also, it can be observed that for $\underset{n\to \infty }{\lim }{a}_n=0$:

$\begin{array}{l}\underset{n\to \infty }{\lim }\frac{{(-1)}^n}{{\left(n\right)}^2}=0\\ \underset{n\to \infty }{\lim }\frac{{(-1)}^{\infty }}{{\left(\infty \right)}^2}\approx 0\end{array}$

Since the denominator is approaching to infinity, so the overall term will approach to zero. The second property is also satisfied.

Therefore, it is proved that the series converges.