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I want to prove the divergence of the following series: $$\sum_{n=1}^{\infty} \frac{1}{n + \ln^2{n}}$$

At first, I tried to find another series who is always smaller to be able to prove that the series diverges by the comparison test. $$\frac{1}{n+\ln^2{n}} > \frac{1}{n+n^2} > \frac{1}{2n^2}$$

But, the resulting series $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges. And I know that the series $\sum_{n=1}^{\infty} \frac{1}{n + \ln^2{n}}$ diverges.

Where my reasoning went wrong!?

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  • $\begingroup$ Is $x$ a constant? $\endgroup$
    – lulu
    Oct 26 '15 at 20:17
  • $\begingroup$ @lulu My mistake. Thank you. $\endgroup$
    – hlapointe
    Oct 26 '15 at 20:18
  • $\begingroup$ No problem. Hint: for large $n$, $log(n)≤\sqrt n$. $\endgroup$
    – lulu
    Oct 26 '15 at 20:19
  • $\begingroup$ @lulu So, rather than replace $\ln^2{n}$ by $n^2$, I should replace it by $\sqrt{n}$? $\endgroup$
    – hlapointe
    Oct 26 '15 at 20:20
  • $\begingroup$ You probably mean $\ln^2 n$. I might use limit comparison with $\sum \frac{1}{n}$. What went wrong is that you gave too much away. From a series is bigger than a converging series we cannot conclude anything. $\endgroup$ Oct 26 '15 at 20:21
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One may observe that, as $n \to \infty$,

$$ \frac{1}{n + \ln^2{n}}\sim \frac{1}{n}\times\frac{1}{1 + \frac{\ln^2{n}}n} \sim \frac1n $$

giving, by using comparison test, the divergence of your initial series.

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    $\begingroup$ I didn't thnik about that. Thanks! $\endgroup$
    – hlapointe
    Oct 26 '15 at 20:27
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For $n\geq 5$, $\ln n \leq \sqrt n$ and hence we have $\frac {1}{(\ln n)^2 +n}\geq \frac{1}{(\sqrt{n})^2+n}=\frac{1}{2n}.$ This implies the divergence of the original series.

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$$\frac{1}{n+\ln^2 n} > \frac{1}{an}$$ iff $$n+\ln^2 n < a n$$ iff $$\ln^2 n < (a-1)n$$ which, for sufficiently large $n$, is true if only $a>1$.

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For any $\alpha>0$ we have $\log x\le \frac{x^{\alpha}}{\alpha}$. Therefore, with $\alpha =1/2$ we have

$$\frac{1}{n+\log^2n}\ge\frac{1}{n+4n}=\frac1{5n}$$

And we are done!

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  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. - Mark $\endgroup$
    – Mark Viola
    Nov 22 '15 at 6:23

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