Analyze if the serie is convergent or divergent $\sum_{n=1}^{\infty}\frac{\ln n}{1+n^2}$

Question

Analyze if the serie is convergent or divergent:

$$\sum_{n=1}^{\infty}\frac{\ln n }{1+n^2}.$$

How do I analyze this?

Is posible comparation? with that series?

Answer 1

HINT

Let consider the convergent

$$\sum_{n=1}^{\infty}\frac{1 }{n^{3/2}}$$

and note that

$$\frac{\frac{\ln n }{1+n^2}}{\frac{1 }{n^{3/2}}}\to 0$$

then refer to limit comparison test.

Answer 2

Consider that $$a_{n}=\frac{ln(n)}{1+n^2}$$, and , $$a_{n+1}=\frac{ln(n+1)}{1+(n+1)^2}$$.

If the limit $$\lim_{n\to\infty}{\frac{a_{n+1}}{a_{n}}}=0$$ , the serie indeed converge . We can further observe that :

$$\lim_{n\to\infty}{\frac{a_{n+1}}{a_{n}}}=\lim_{n\to\infty}\frac{\frac{ln(n+1)}{1+(n+1)^2}}{\frac{ln(n)}{1+n^2}} \rightarrow 1 \neq0$$ , and hence is divergent.

Hint:

Computing the limit is trivial and left to the reader.

Answer 3

Make use of the following:

\begin{align} \sum_{n=1}^{\infty} \frac{\log n}{1+n^2}<\sum_{n=1}^{\infty} \frac{\log n}{n^2}=-\zeta'(2)=0.9375... \end{align}