How do you use the limit comparison test to determine if $\sum{\dfrac{n}{{{n}^{2}}+1}}$ from $\left[ 1,\infty \right)$ is convergent or divergent?

Hint: In this problem we need to check whether the given series is convergence or divergence in the given limits by using the limit comparison test. For a limit comparison test we need to have two series let’s say ${{a}_{n}}$, ${{b}_{n}}$ such that ${{a}_{n}}\ge 0$, ${{b}_{n}}>0$. If we know whether the series ${{b}_{n}}$ divergence or convergence from the value of $\displaystyle \lim_{n \to \infty }\dfrac{{{a}_{n}}}{{{b}_{n}}}$ we can decide whether the series ${{a}_{n}}$ divergence or convergence. In this problem we have only one series which is $\sum{\dfrac{n}{{{n}^{2}}+1}}$, so we will assume this series as ${{a}_{n}}$. For another series which is ${{b}_{n}}$ we will consider another series which is greater than the given series. After that we will do the limit comparison test and test whether the given function convergence or divergence.

Complete step by step answer:
The series is $\sum{\dfrac{n}{{{n}^{2}}+1}}$.
Given limits are $\left[ 1,\infty \right)$.
In the problem they have mentioned to use the limit comparison test. So, assume the given series as
$\Rightarrow {{a}_{n}}=\sum{\dfrac{n}{{{n}^{2}}+1}}$
For limit comparison tests we need to have another series which is greater than the given series. So, we are going to assume another series as
$\Rightarrow {{b}_{n}}=\sum{\dfrac{1}{n}}$
Performing the limit comparison test, then we will get
$\begin{align}
  & \Rightarrow \displaystyle \lim_{n \to \infty }\dfrac{{{a}_{n}}}{{{b}_{n}}}=\displaystyle \lim_{n \to \infty }\dfrac{\dfrac{n}{{{n}^{2}}+1}}{\dfrac{1}{n}} \\
 & \Rightarrow \displaystyle \lim_{n \to \infty }\dfrac{{{a}_{n}}}{{{b}_{n}}}=\displaystyle \lim_{n \to \infty }\dfrac{n}{{{n}^{2}}+1}\times \dfrac{n}{1} \\
 & \Rightarrow \displaystyle \lim_{n \to \infty }\dfrac{{{a}_{n}}}{{{b}_{n}}}=\displaystyle \lim_{n \to \infty }\dfrac{{{n}^{2}}}{{{n}^{2}}\left( 1+\dfrac{1}{{{n}^{2}}} \right)} \\
 & \Rightarrow \displaystyle \lim_{n \to \infty }\dfrac{{{a}_{n}}}{{{b}_{n}}}=\displaystyle \lim_{n \to \infty }\dfrac{1}{1+\dfrac{1}{{{n}^{2}}}} \\
\end{align}$
Here we have the limit value as positive finite.
We know that the considered series which is ${{b}_{n}}=\sum{\dfrac{1}{n}}$. Hence the series ${{a}_{n}}=\sum{\dfrac{n}{{{n}^{2}}+1}}$ also divergence in the given limits.

Note:
In this problem we have the limit value as the positive finite and the considered series diverges, so we have written that the given series also diverges. If the considered series converges then the given function also converges.