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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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4 Answers 4

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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

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

See related techniques.

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  • $\begingroup$ 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? $\endgroup$
    – Nick
    Commented Mar 22, 2014 at 17:43
  • $\begingroup$ @Nick: $\frac{1}{ne^{n}}\leq \frac{1}{e^n}$. $\endgroup$ Commented Mar 22, 2014 at 18:11
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$1+e^{-n}>e^{-n}\Rightarrow \ln (1+e^{-n})> \ln e^{-n}=-n$

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

What can you conclude?

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  • $\begingroup$ $\dfrac {\ln(1+e^{-n})}{n}> -1$ and the origianl series is greater then the series of -1 which is divergent. Am I right? $\endgroup$
    – Nick
    Commented Mar 22, 2014 at 18:32
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    $\begingroup$ 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... $\endgroup$
    – DonAntonio
    Commented Mar 22, 2014 at 18:36
  • $\begingroup$ @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.... $\endgroup$
    – user87543
    Commented Mar 23, 2014 at 3:58
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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

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  • $\begingroup$ This answer has an error. Simply check [n=2,log(1+exp(-n))] \\ = [2, 0.126928011043], [n=10,log(1+exp(-n))] \\ = [10, 0.0000453988992169] [n=100,log(1+exp(-n))] \\ = [100, 3.72007597602 E-44] $\endgroup$ Commented Oct 16, 2021 at 7:45
  • $\begingroup$ Removed the downvoting due to correction of the long-time ignored error in formula. $\endgroup$ Commented Oct 22, 2021 at 7:22

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