Test the following series for convergence

$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{0}}^{{}^{\mathbf{\infty }}}\frac{\mathbf{(}\mathbf{-}\mathbf{1}{\mathbf{)}}^{\mathbf{n}}\mathbf{n}}{\mathbf{1}\mathbf{+}{\mathbf{n}}^{\mathbf{2}}}$

If the terms' absolute values gradually decline to zero, or if$\mathbf{\left|}{\mathit{a}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{\right|}\mathbf{\le }\mathbf{\left|}{\mathit{a}}_{\mathbf{n}}\mathbf{\right|}$and$\underset{{}_{\mathbf{n}\mathbf{\to }\mathbf{\infty }}}{\mathbf{l}\mathbf{i}\mathbf{m}}{\mathit{a}}_{\mathbf{n}}\mathbf{=}\mathbf{0}\mathbf{.}$, then an alternating series converges.

The alternating series is$\sum _{\mathrm{n}=0}^{\infty }\frac{(-1{)}^{\mathrm{n}}\mathrm{n}}{1+{\mathrm{n}}^2}$.

Let test the series for convergence by using the following test

$\mathbf{\left|}{\mathit{a}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{\right|}\mathbf{\le }\mathbf{\left|}{\mathit{a}}_{\mathbf{n}}\mathbf{\right|}$and $\mathit{l}\mathit{i}{\mathit{m}}_{\mathbf{n}\mathbf{\to }\mathbf{\infty }}{\mathit{a}}_{\mathbf{n}}\mathbf{=}\mathbf{0}\mathbf{.}$

The series converges if satisfies the condition. So,

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

Now check if$\left|a_{\mathrm{n}+1}\right|\le \left|a_{\mathrm{n}}\right|$using the following method.

Forlocalid="1658727024584" $\frac{|a_n+1|}{\left|a_n\right|}<1$this means$\left|a_{\mathrm{n}+1}\right|\le \left|a_{\mathrm{n}}\right|$.

Forlocalid="1658726433620" $\frac{|a_n+1|}{\left|a_n\right|}>1$this means$\left|a_{\mathrm{n}+1}\right|\le \left|a_{\mathrm{n}}\right|$

Using the method for the series get:

$$\begin{gathered} \frac{|a_n+1|}{\left|a_n\right|}=\frac{{\displaystyle \frac{n+1}{1+{\left(n+1\right)}^2}}}{{\displaystyle \frac{n}{n+1^2}}} \\ \frac{|a_n+1|}{\left|a_n\right|}=\frac{\left(n+1\right)\left(1+n^2\right)}{n\left(1+{\left(1+n\right)}^2\right)} \\ =\frac{n^3+n^2+n+1}{n^3+2n^2+2n} \end{gathered}$$

From the expression from the previous step,

If $\mathbf{(}{\mathit{n}}^{\mathbf{3}}\mathbf{+}{\mathit{n}}^{\mathbf{2}}\mathbf{+}\mathit{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{-}\mathbf{(}{\mathit{n}}^{\mathbf{3}}\mathbf{+}\mathbf{2}{\mathit{n}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathit{n}\mathbf{)}\mathbf{<}\mathbf{0}$then $\frac{{\mathbf{n}}^{\mathbf{3}}\mathbf{+}{\mathbf{n}}^{\mathbf{2}}\mathbf{+}\mathbf{n}\mathbf{+}\mathbf{1}}{{\mathbf{n}}^{\mathbf{3}}\mathbf{+}\mathbf{2}{\mathbf{n}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{n}}\mathbf{<}\mathbf{1}$

role="math" localid="1658726957852" $(n^3+n^2+n+1)-(n^3+2n^2+2n)<0=1-n^2-n$

When $n\ge 1,1-n^2-n<0$

Hence,

role="math" localid="1658727082841" $\frac{|a_n+1|}{\left|a_n\right|}<1,|a_n+1)|<|a_n|$

$\underset{\mathrm{n}\to \infty }{lim}{a}_{\mathrm{n}}=0=\underset{\mathrm{n}\to \infty }{\lim }\frac{n}{1+n^2}$

Hence, the series $\sum _{\mathrm{n}=0}^{\infty }\frac{(-1{)}^{\mathrm{n}}\mathrm{n}}{1+{\mathrm{n}}^2}$is, converges