Does the series Σln(1+e^-n)/n converge?

lukatwo
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Homework Statement



So I need to determine if the series \Sigmaln(1+e^{-n})/n converges.

Homework Equations


The Attempt at a Solution



I know it does, but cannot prove it. Wolfram says that the ratio test indicates that the series converges, but when I try to solve the limit I get that it equals 1(which is not conclusive). here
 
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Do you mean
\sum \frac{ln(1+ e^n}{n}
 
lukatwo said:

Homework Statement



So I need to determine if the series \Sigmaln(1+e^-n)/n converges.


Homework Equations





The Attempt at a Solution



I know it does, but cannot prove it. Wolfram says that the ratio test indicates that the series converges, but when I try to solve the limit I get that it equals 1(which is not conclusive). here

The series you posted looks like ln(1+e^(-n))/n. The series you tested in Mathematics looks like ln(1+e^(1/n))/n. e^(-n) is pretty different from e^(1/n).
 
No it's -n alright, but I've been switching them up along the way. Now I see my problem. Thanks
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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