# Proving divergence by using comparison test

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!?

• Is $x$ a constant?
– lulu
Oct 26 '15 at 20:17
• @lulu My mistake. Thank you. Oct 26 '15 at 20:18
• No problem. Hint: for large $n$, $log(n)≤\sqrt n$.
– lulu
Oct 26 '15 at 20:19
• @lulu So, rather than replace $\ln^2{n}$ by $n^2$, I should replace it by $\sqrt{n}$? Oct 26 '15 at 20:20
• 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. Oct 26 '15 at 20:21

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.

• I didn't thnik about that. Thanks! Oct 26 '15 at 20:27

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.

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

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!

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