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$\lim_{t\to1^-}(1-t)\sum_{r=1}^{\infty}\frac{t^r}{1+t^r}$

My approach $\frac{t^r}{1+t^r}=t^r-t^{2r}+t^{3r}-\cdots$ $\implies \sum_{r=1}^{\infty}\frac{t^r}{1+t^r}=\frac{t}{1-t}-\frac{t^2}{1-t^2}+\frac{t^3}{1-t^3}-\cdots$ $\implies(1-t)\sum_{r=1}^{\infty}\frac{t^r}{1+t^r}=t-\frac{t^2}{1+t}+\frac{t^3}{1+t+t^2}-\cdots$ $\implies \lim_{t\to1^-}(1-t)\sum_{r=1}^{\infty}\frac{t^r}{1+t^r}=1-\frac12+\frac13-\frac14+\cdots=\ln 2$

Is this correct? Any other more rigorous approach ?

Solution provided by problem poser:

$\lim_{t\to1^-}(1-t)\sum_{r=1}^{\infty}\frac{t^r}{1+t^r}=\lim_{t\to1^-}-\ln t\sum_{r=1}^{\infty}\frac{1}{1+e^{-r\ln t}}$ $=\lim_{n\to\infty}\frac1n\sum_{r=1}^{\infty}\frac{1}{1+e^{\frac rn}}$ $=\int_{0}^{1}\frac{1}{1+e^x}dx=\ln\frac {2e}{1+e}$

This answer seems to be incorrect as

$\frac{t^r}{2}<\frac{t^r}{1+t^r}<t^r\implies \frac{t}{2(1-t)}\le\sum_{r=1}^{\infty}\frac{t^r}{1+t^r}\le\frac{t}{1-t}\implies\frac12\le\lim_{t\to1^-}(1-t)\sum_{r=1}^{\infty}\frac{t^r}{1+t^r}\le 1.$

But $\ln\frac{2e}{1+e}\approx 0.38$

Paramanand Singh
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Makar
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    Also your answer $\ln 2$ and approach is correct. Solution by integrals has an error. The desired integral is $\int_{0}^{\infty}\frac{dx}{1+e^x}=\log 2$. – Paramanand Singh Apr 16 '19 at 01:41
  • One way would be to show that $$\lim_{t \uparrow 1} (1 - t) \sum_{r \geq 1} \frac {t^r} {1 + t^r} = \lim_{t \uparrow 1} (1 - t) \int_1^\infty \frac {t^r} {1 + t^r} dt = -\lim_{t \uparrow 1} \frac {(1 - t) \ln(1 + t)} {\ln t}.$$ Or to show that the limits can be interchanged in $$\lim_{n \to \infty} \frac 1 n \sum_{r \geq 1} \frac 1 {1 + e^{r/n}} = \lim_{n \to \infty} \lim_{A \to \infty} \frac 1 n \sum_{1 \leq r \leq A n} \frac 1 {1 + e^{r/n}} = \lim_{A \to \infty} \int_0^A \frac {dt} {1 + e^t}.$$ – Maxim Apr 16 '19 at 11:35
  • @ParamanandSingh Do you mean the line with the inequalities? Why is it wrong? – Maxim Apr 16 '19 at 11:37
  • @Maxim: well those inequalities are fine. I have deleted my comment about that. – Paramanand Singh Apr 16 '19 at 12:13
  • Can someone explain the second step of the solution given by the problem poser? I cannot see where it came from. – Tony Apr 18 '19 at 05:45
  • @Tony: Note that $(1-t)=\dfrac{1-t}{-\ln t} \cdot (-\ln t) $ and the first factor tends to $1$ so all that remains is $-\ln t$. Further one sets $-\ln t=1/n$ so that $n\to\infty $. – Paramanand Singh Apr 18 '19 at 07:13

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