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.

• Do you really mean $x\to-1$? Apr 17, 2019 at 16:12
• Should surely be $t \to -1$ Apr 17, 2019 at 16:12
• edited the question. Does partial fraction help? Apr 17, 2019 at 16:16
• Do you really mean $t\to-1$? Apr 17, 2019 at 16:31
• See the same question asked earlier: math.stackexchange.com/q/3187497/72031 Apr 18, 2019 at 3:26

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}

• Did that, but didn't find anything remarkable. :(
– Tony
Apr 17, 2019 at 17:13
• @Tony Ok, I've added athe full solution. Apr 17, 2019 at 20:00
• But the answer given is $\ln \big(\frac{2e}{1+e}\big)$
– Tony
Apr 18, 2019 at 1:31
• Funny, that's where I ended up too, once I realized my error in sign. Might try a quick computation to see where it's headed. Apr 18, 2019 at 2:19
• @BrianTung: The given answer is wrong. See some discussion on this at math.stackexchange.com/q/3187497/72031 Apr 18, 2019 at 3:27