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:

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

score 12 · Accepted Answer · answered Mar 22 '14 at 17:19

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 $x\gt 0$.

score 2 · Answer 2 · edited Apr 13 '17 at 12:21

2

For large $n$, we have

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

See related techniques.

edited Apr 13 '17 at 12:21

Community

1

answered Mar 22 '14 at 17:32

Mhenni Benghorbal

47,431

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

score 0 · Answer 3 · answered Mar 22 '14 at 17:26

0

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

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

What can you conclude?

answered Mar 22 '14 at 17:26

$\dfrac {\ln(1+e^{-n})}{n}> -1$ and the origianl series is greater then the series of -1 which is divergent. Am I right? – Nick Mar 22 '14 at 18:32
1

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.... – Mar 23 '14 at 03:58

malloc · Answer 4 · 2021-10-16T21:46:58.153

0

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

edited Oct 16 '21 at 21:46

answered Mar 22 '14 at 20:31

malloc

407

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] – Gottfried Helms Oct 16 '21 at 07:45
Removed the downvoting due to correction of the long-time ignored error in formula. – Gottfried Helms Oct 22 '21 at 07:22

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

4 Answers4