Does $\sum_{n=1}^{\infty} \frac{\ln(n)}{n^2+2}$ converge? I'm trying to find out whether
$$\sum_{n=1}^{\infty} \frac{\ln(n)}{n^2+2}$$
is convergent or divergent?
 A: Let $s \in (0,1)$. Since $\displaystyle \lim_{n \to\infty}n^{-s}\ln n=0$, there is an integer $k=k(s) \in \mathbb{N}$ such that
$$
n^{-s}\ln n \le 1 \quad \forall n > k.
$$
It follows that
$$
\sum_{n=1}^\infty\frac{\ln n}{n^2+2}\le\sum_{n=1}^{k}\frac{\ln n}{n^2+2}+\sum_{n>k}\frac{1}{n^{2-s}}\le \sum_{n=1}^{k}\frac{\ln n}{n^2+2}+\sum_{n=1}^\infty\frac{1}{n^{2-s}}<\infty.
$$
A: Clearly 
$$
\frac{\ln(n)}{n^2+2}\leq \frac{\ln(n)}{n^2}
$$
You can apply the integral test to show that $\sum\frac{\ln(n)}{n^2}$ converges.  You only need to check that $\frac{\ln(n)}{n^2}$ is decreasing.  But, the derivative is clearly negative for $n>e$.
A: $$\sum_{n=1}^{\infty}{\frac{\ln(n)}{n^2+2}}$$
clearly
$$\sum_{n=1}^{\infty}{\frac{\ln(n)}{n^2+2}}<\sum_{n=1}^{\infty}{\frac{\ln(n)}{n^2}}$$
We have
$$\ln{n}<\sqrt{n}$$
because
$$\lim_{n\rightarrow \infty}\frac{\ln{n}}{\sqrt{n}}=\lim_{n\rightarrow\infty}\frac{\frac{1}{n}}{\frac{1}{2\sqrt{n}}}=\lim_{n\rightarrow\infty}\frac{2\sqrt{n}}{n}=\lim_{n\rightarrow\infty}\frac{2n^{1/2}}{n}=\lim_{n\rightarrow\infty}\frac{2}{n^{1/2}}=0$$
as
$$\ln{n}<\sqrt{n}\rightarrow\frac{\ln{n}}{n^{2}}<\frac{\sqrt{n}}{n^{2}}=\frac{n^{1/2}}{n^2}=\frac{1}{n^{3/2}}$$
Soon
Therefore, we have $\sum_{n=1}^{\infty}{\frac{1}{n^{3/2}}}$ is a hyper harmonic series (p-series) with $ p> 1 $, hence is convergent.
By comparison test and by
$$\sum_{n=1}^{\infty}{\frac{\ln{n}}{n^2+2}}<\sum_{n=1}^{\infty}{\frac{\ln{n}}{n^2}}<\sum_{n=1}^{\infty}{\frac{1}{n^{3/2}}}$$
We have
$$\sum_{n=1}^{\infty}{\frac{\ln{n}}{n^2+2}}$$ is convergent serie.
A: The series can be estimated termwise as follows:
\begin{eqnarray}
\frac{\ln(n)}{n^2 + 2} & \leq & \frac{\ln(n)}{n^2} = \ln(n) e^{-2\ln(n)} = 2 \cdot \frac{1}{2} \ln(n) e^{-2\ln(n)} \leq 2 \cdot \bigg[ \sum_{k=0}^\infty \frac{1}{k!}\bigg(\frac{1}{2} \ln(n)\bigg)^k \bigg] e^{-2\ln(n)} \\
& = & 2 \cdot e^{\frac{1}{2}\ln(n)} e^{-2\ln(n)} = 2 \cdot e^{-\frac{3}{2} \ln(n)} = 2 \cdot n^{-3/2} \ .
\end{eqnarray}
Now the series converges because it is non-negative and has a converging majorant.
A: $\forall \alpha,\beta >0, \exists x_0 \in \mathbb{R}^+$ such that $\forall x>x_0$:
$$(\ln x)^\alpha < x^\beta$$
Setting $\alpha=1,\beta=\frac{1}{2} $ will prove that the sum is convergent.
A: A related problem. Just note this
$$ a_n=\frac{\ln(n)}{n^2+2}\sim \frac{\ln(n)}{n^2}=b_n. $$
Now, 
$$ \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{n^2}{n^2+2}=1.$$
So, the two series converge together or diverge together and since $\sum_{n}b_n$ converges (integral test), then so does $\sum_{n}a_n$.  
