$$\lim_{t\to 1^{-}}(1-t)\sum_{r = 1}^\infty \frac{t^r}{t^r+1}$$
Note: I am a high school student and this problem appeared in my test. So, please try to use methods to solve this problem at a high school level :)
My Attempt:
I have honestly no idea how to approach this problem.
I first tried to simplify the summation but didn't find any pattern. By looking at the options given to me ( which were all in ln's and e's )I do get a feel that we may have to integrate at some point. Though I am not sure.
Any help would be appreciated.
Hint: write $\frac{1}{1+t^r}$ using the formula for a sum of a geometric series, and change the order of summation.
Full solution: \begin{align} \lim_{t\rightarrow 1^-} (1-t) \sum_{r=1}^\infty \frac{t^r}{1+t^r} &= \lim_{t\rightarrow 1^-} (1-t) \sum_{r=1}^\infty t^r \sum_{n=0}^\infty (-t^r)^n = \\ &= \lim_{t\rightarrow 1^-} (1-t) \sum_{n=0}^\infty \sum_{r=1}^\infty (-1)^n t^{(n+1)r} = \\ &= \lim_{t\rightarrow 1^-} (1-t) \sum_{n=0}^\infty (-1)^n\frac{t^{n+1}}{1-t^{n+1}} = \\ &= \lim_{t\rightarrow 1^-} \sum_{n=0}^\infty (-1)^n \frac{1 -t}{1-t^{n+1}} t^{n+1} = \\ &= \lim_{t\rightarrow 1^-} \sum_{n=0}^\infty (-1)^n \frac{1}{1+t+t^2+\dots+t^n} t^{n+1} = \\ &= \sum_{n=0}^\infty (-1)^n \frac{1}{n+1} = \\ &= \ln 2\end{align}