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I'm having difficulties proving the divergence of the series

$$\sum_{n=1}^\infty\frac{1}{\log(1+e^n)}.$$

Wolfram|Alpha suggest to use the Comparison test. I tried to look for a lower estimate to force it up to infinity but I couldn't come up with something useful.

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Writing $1+e^n=e^n(1+e^{-n})$, we have $$ \log(1+e^n)=n+\log(1+e^{-n})\leq n+\log 2\leq 2n $$ for all $n\geq 1$. Therefore $$ \frac{1}{\log(1+e^n)}\geq \frac{1}{2n} $$ and hence $\sum_{n=1}^{\infty}\frac{1}{\log(1+e^n)}$ diverges by comparison with the harmonic series.

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As it is a series with positive terms, you can use equivalents and it will be very short: \begin{alignat}{3}1+\mathrm e^n&\sim_\infty\mathrm e^n,\enspace\text{hence}&\qquad &&\frac1{\log(1+\mathrm e^n)}\sim_\infty \frac1{\log \mathrm e^n}=\frac1n,&\qquad\qquad&\end{alignat} which diverges.

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