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

Question

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

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.

@HAMIDINESOUMARE: $r$ is just a dummy variable (the index of summation). I don't see what else can be said about it. (Other than it ranges over all positive integers.) — Brian Tung, Apr 17 '19 at 16:42
Well, I ended up at $-\ln 2$, once I realized I had made an error in sign. I wonder what's going on here. As I said in my comment to Tony's answer, I might try to do a quick computation when I get a chance. — Brian Tung, Apr 18 '19 at 02:20
See the same question asked earlier: https://math.stackexchange.com/q/3187497/72031 — Paramanand Singh, Apr 18 '19 at 03:26

Adam Latosiński · Accepted Answer · 2019-04-18T08:23:04.453

10

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}

edited Apr 18 '19 at 08:23

answered Apr 17 '19 at 17:02

Adam Latosiński

10,558

Did that, but didn't find anything remarkable. :( – Tony Apr 17 '19 at 17:13
@Tony Ok, I've added athe full solution. – Adam Latosiński Apr 17 '19 at 20:00
But the answer given is $\ln \big(\frac{2e}{1+e}\big)$ – Tony Apr 18 '19 at 01:31
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. – Brian Tung Apr 18 '19 at 02:19
Oppss... There is a typo in the question. I have updated it. Sorry... My bad! – Tony Apr 18 '19 at 02:23
That still only changes the sign back to my answer (which otherwise matches Adam's), and my computation gives $0.69305\ldots$, which is very close to $\ln 2$. Unless there's a further error in the question, I think the given answer might be wrong. – Brian Tung Apr 18 '19 at 02:27
1

@BrianTung: The given answer is wrong. See some discussion on this at https://math.stackexchange.com/q/3187497/72031 – Paramanand Singh Apr 18 '19 at 03:27
@BrianTung The question is as it is copied from my book. There's no further typo in that. I think the given answer might be incorrect. – Tony Apr 18 '19 at 05:38
@Adam I suggest you edit your answer as per the correct question. – Tony Apr 18 '19 at 06:04
@ParamanandSingh: Interesting! – Brian Tung Apr 18 '19 at 07:08
Corrected the answer to fit the corrected question. – Adam Latosiński Apr 18 '19 at 08:23

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

1 Answers1