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

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

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}
• But the answer given is $\ln \big(\frac{2e}{1+e}\big)$ – Tony Apr 18 at 1:31