91Ó°ÊÓ

If the partial sums${\mathbf{\text{S}}}_{\mathbf{n}}$ of an infinite series tend to a limit S, the series is called convergent. If the partial sums${\mathbf{\text{S}}}_{\mathbf{n}}$ of an infinite series don't approach a limit, the series is called divergent.

The limiting value S is called the sum of the series.

The given series is $\underset{1}{\overset{∞}{∑}}\frac{n}{{\left(n^2+1\right)}^2}$.

Use the integral in the given series, $\underset{}{\overset{∞}{∫}}\frac{n}{{\left(n^2+1\right)}^2} dn$.

Now solve the integral is as follows:

Let $t=n^2+1$,$dt=2n dn$

Substitute the values of t and dt into the integral and change the variable n from into t.

$\underset{}{\overset{∞}{∫}}\frac{dt}{2t^2}=\underset{}{\overset{∞}{∫}}\frac{1}{2} t^{-2} dt$

Use the integral formula, $∫x^a dx=\frac{x^{a+1}}{a+1} dx+c$, where is a constant.

$\begin{array}{rcl}\underset{}{\overset{∞}{∫}}\frac{1}{2} t^{-2} dt& =& \frac{1}{2}{\left[\frac{t^{-2+1}}{-2+1}\right]}^{∞}\\ & =& \frac{1}{2}{\left[\frac{t^{-1}}{-1}\right]}^{∞}\\ & =& -\frac{1}{2}\left[{∞}^{-1}\right]\\ & =& ∞\\ & & \end{array}$

Hence, the series approaches to infinite therefore the given series diverges.