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

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.

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  • $\begingroup$ Do you really mean $x\to-1$? $\endgroup$ Apr 17, 2019 at 16:12
  • $\begingroup$ Should surely be $t \to -1$ $\endgroup$
    – fGDu94
    Apr 17, 2019 at 16:12
  • $\begingroup$ edited the question. Does partial fraction help? $\endgroup$
    – vidyarthi
    Apr 17, 2019 at 16:16
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    $\begingroup$ Do you really mean $t\to-1$? $\endgroup$ Apr 17, 2019 at 16:31
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    $\begingroup$ See the same question asked earlier: math.stackexchange.com/q/3187497/72031 $\endgroup$
    – Paramanand Singh
    Apr 18, 2019 at 3:26

1 Answer 1

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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}

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  • $\begingroup$ Did that, but didn't find anything remarkable. :( $\endgroup$
    – Tony
    Apr 17, 2019 at 17:13
  • $\begingroup$ @Tony Ok, I've added athe full solution. $\endgroup$ Apr 17, 2019 at 20:00
  • $\begingroup$ But the answer given is $\ln \big(\frac{2e}{1+e}\big)$ $\endgroup$
    – Tony
    Apr 18, 2019 at 1:31
  • $\begingroup$ 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. $\endgroup$
    – Brian Tung
    Apr 18, 2019 at 2:19
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    $\begingroup$ @BrianTung: The given answer is wrong. See some discussion on this at math.stackexchange.com/q/3187497/72031 $\endgroup$
    – Paramanand Singh
    Apr 18, 2019 at 3:27

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