Question

Test the following series for convergence.

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

Short answer

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

Step-by-step solution

Given information

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

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

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

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

To check use the following method:

If $\frac{\left|a_{n+1}\right|}{\left|a_n\right|}<1$, that means $\left|a_n\right|>\left|a_{n+1}\right|$.

And, if $\frac{\left|a_{n+1}\right|}{\left|a_n\right|}>1$, that means $\left|a_n\right|<\left|a_{n+1}\right|$. Thus:

$\begin{array}{c}\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=\frac{\frac{{\left(2\right)}^{n+1}}{{(n+1)}^2}}{\frac{{\left(2\right)}^n}{n^2}}\\ =\frac{n^2{\left(2\right)}^{n+1}}{{(n+1)}^2{\left(2\right)}^n}\\ =\frac{2n^2}{n^2+2n+1}\end{array}$

If $\left(2n^2\right)-\left(n^2+2n+1\right)<0$, then $\frac{2n^2}{n^2+2n+1}<1$.

$\left(2n^2\right)-\left(n^2+2n+1\right)=n^2-2n-1$

For $n\ge 3$, $n^2-2n-1>0$.

That means $\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=\frac{\frac{{\left(2\right)}^{n+1}}{{(n+1)}^2}}{\frac{{\left(2\right)}^n}{n^2}}>1$for $n\ge 3$.

Since, the first property is not satisfied, so the series diverges.