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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?

Gibbs
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3 Answers3

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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.

gt6989b
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user
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  • In what sense is this answer a "hint"? – Najib Idrissi Jun 06 '18 at 22:15
  • @NajibIdrissi I understand your observation, it is a strong hint but it let the final work to the OP in order to the application of the limit comparison test for the case with limit $\to 0$. – user Jun 06 '18 at 22:18
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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}

Diffusion
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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.

Noodle
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    Are you sure that $$\lim_{n\to\infty}{\frac{a_{n+1}}{a_{n}}}=\lim_{n\to\infty}\frac{\frac{\ln(n+1)}{1+(n+1)^2}}{\frac{\ln(n)}{1+n^2}} \rightarrow 0$$ – user Jun 06 '18 at 22:37
  • Looks like I made a mistake computing it , indeed it is 1 and the series diverge , thanks for noticing . – Noodle Jun 06 '18 at 23:11
  • If the results of the ratio-test is 1, you cannot make any conclusion about the divergence of convergence of your series; the test is then inconclusive. Here are some extensions of the ratio-test that you can use in such a case: https://en.wikipedia.org/wiki/Ratio_test – Diffusion Jun 06 '18 at 23:57