# Convergent or divergent

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

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

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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 $x\gt 0$.

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For large $n$, we have

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

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Can I say: $\sum\limits_{n=1}^{\infty} \dfrac {ln(1+e^{-n})}{n} \sim \sum\limits_{n=1}^{\infty} \dfrac {e^{-n}}{n}$ and now use d'Alembert's ratio test? –  Nick Mar 22 '14 at 17:43
@Nick: $\frac{1}{ne^{n}}\leq \frac{1}{e^n}$. –  Mhenni Benghorbal Mar 22 '14 at 18:11

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

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

What can you conclude?

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$\dfrac {\ln(1+e^{-n})}{n}> -1$ and the origianl series is greater then the series of -1 which is divergent. Am I right? –  Nick Mar 22 '14 at 18:32
I don't think anything interesting about convergency can be concluded, perhaps because the inequalities are trivial and yield negative terms on one of their sides... –  DonAntonio Mar 22 '14 at 18:36
@DonAntonio : I was hoping that $\dfrac {\ln(1+e^{-n})}{n}> -1$ implies $\sum_{n=1}^{\infty} \dfrac {\ln(1+e^{-n})}{n}> \sum_{n=1}^{\infty} -1$ left hand side is divergent so would be right hand side.. then i realized comparision test is valid only for positive terms.... I realize my mistake.... –  Praphulla Koushik Mar 23 '14 at 3:58

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 $\ln(e^{-n})$ and with properties of logarithms you get $-nln(e)=-n$ and : $$ln(1+e^{-n}) \sim-n$$ $$\sum\limits_{n=1}^{\infty} \dfrac {\ln(1+e^{-n})}{n} \sim \sum\limits_{n=1}^{\infty} \dfrac {-n}{n}.$$ So..try to continue by yourself

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