Convergence of $\sum\limits_{n=1}^{\infty} \frac {\ln(1+e^{-n})}{n}. $

Question

For homework (Calculus 2) I have to determine does this series converge or diverge and I don't know how to start:

$$\sum\limits_{n=1}^{\infty} \dfrac {\ln(1+e^{-n})}{n}. $$

Answer 1

Hint: For $x\gt 0$, we have $\ln(1+x)\lt x$.

Remark: This inequality has many proofs. One way is to exponentiate. We get the equivalent inequality $1+x\lt e^x$, which is clear from the power series of $e^x$. Or else we can let $f(x)=x-\ln(1+x)$. Note that $f(0)=0$ and $f'(x)\gt 0$ when $x\gt 0$.

Answer 2

$1+e^{-n}>e^{-n}\Rightarrow \ln (1+e^{-n})> \ln e^{-n}=-n$

So, $\dfrac {\ln(1+e^{-n})}{n}> ??$

What can you conclude?

Answer 3

For large $n$, we have

$$ \ln(1+e^{-n}) \sim e^{-n}. $$

See related techniques.

Answer 4

When you have this kind of series delete the constant value and try to apply some basic property. For the firsts exercises this should be enough.

You can approximate the $\ln(1+e^{-n})$ to $ e^{-n}$ and :

$$\sum\limits_{n=1}^{\infty} \dfrac {\ln(1+e^{-n})}{n} \sim \sum\limits_{n=1}^{\infty} \dfrac {e^{-n}}{n}.$$ So..try to continue by yourself