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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!?
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.
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!
