Evaluate $$l=\lim_{t\to1^{-}} (1-t) \sum_{r=1}^\infty\frac{t^r}{1+t^r}$$
My solution:
$$l=\lim_{t\to1^{-}} \frac{(1-t)}{\ln(t)}\cdot \ln(t)\sum_{r=1}^{\infty} \frac1{1+t^{-r}}$$
$$=\lim_{t\to1^{-}}-\ln(t) \sum_{r=1}^{\infty}\frac1{1+e^{-r\ln(t)}}$$
Let $-\ln(t)=\frac1n,$ as $t\to1^{-},\ n\to+\infty$
So $$l=\lim_{n\to+\infty}\frac1n \sum_{r=1}^{\infty}\frac1{1+e^{r/n}}$$
$$=\int_{0}^{1}\frac{dx}{1+e^x}=\ln \left(\frac{2e}{1+e}\right)$$
Is there any other way to do this question?
Best Answer
You can put $t=e^{-h} $ so that $h\to 0^+$ and then $h/(1-t)\to 1$. The expression under limit can thus be reduced to $$h\sum_{r=1}^{\infty} \frac{1}{1+e^{rh}}$$ The function $f(x) =1/(1+e^x)$ is decreasing on $[0,\infty) $ therefore $$\int_{h} ^{(n+1)h}f(x)\,dx\leq h\sum_{r=1}^{n}f(rh) \leq\int_{0}^{nh}f(x)\,dx$$ Letting $n\to\infty $ we get $$\int_{h} ^{\infty} f(x) \, dx\leq h\sum_{r=1}^{\infty} f(rh) \leq \int_{0}^{\infty} f(x) \, dx$$ Letting $h\to 0^{+}$ we get the desired limit as $\int_{0}^{\infty} f(x) \, dx=\log 2$.
The above technique has also been used in this answer.