Evaluate $$l=\lim_{t\to1^{-}} (1-t) \sum_{r=1}^\infty\frac{t^r}{1+t^r}$$

My solution:

$$l=\lim_{t\to1^{-}} \frac{(1-t)}{\ln(t)}\cdot \ln(t)\sum_{r=1}^{\infty} \frac1{1+t^{-r}}$$

$$=\lim_{t\to1^{-}}-\ln(t) \sum_{r=1}^{\infty}\frac1{1+e^{-r\ln(t)}}$$

Let $-\ln(t)=\frac1n,$ as $t\to1^{-},\ n\to+\infty$

So $$l=\lim_{n\to+\infty}\frac1n \sum_{r=1}^{\infty}\frac1{1+e^{r/n}}$$

$$=\int_{0}^{1}\frac{dx}{1+e^x}=\ln \left(\frac{2e}{1+e}\right)$$

Is there any other way to do this question?

  • $\begingroup$ I wonder about a problem somewhere. Compute it for $t=\frac 9{10}$ to get $0.633101$; more serious, $t=\frac {99}{100}$ to get $0.689169$. Do you see where we are going ? $\endgroup$ Commented May 14, 2020 at 11:14
  • $\begingroup$ I just wrote "I wonder". I do not use Desmos and I made summation to infinity. wolframalpha.com/input/… $\endgroup$ Commented May 14, 2020 at 11:33
  • 1
    $\begingroup$ How did you switch to integral from sum? The result you seem to assume holds only when upper limit of sum is $n$ instead of $\infty$. $\endgroup$
    – Paramanand Singh
    Commented May 14, 2020 at 13:50
  • 1
    $\begingroup$ @ClaudeLeibovici: your numerical calculations are correct (as always) and the desired limit is $\log 2$. $\endgroup$
    – Paramanand Singh
    Commented May 14, 2020 at 14:16
  • 1
    $\begingroup$ Don't worry, please. Cheers $\endgroup$ Commented May 14, 2020 at 14:27

2 Answers 2


You can put $t=e^{-h} $ so that $h\to 0^+$ and then $h/(1-t)\to 1$. The expression under limit can thus be reduced to $$h\sum_{r=1}^{\infty} \frac{1}{1+e^{rh}}$$ The function $f(x) =1/(1+e^x)$ is decreasing on $[0,\infty) $ therefore $$\int_{h} ^{(n+1)h}f(x)\,dx\leq h\sum_{r=1}^{n}f(rh) \leq\int_{0}^{nh}f(x)\,dx$$ Letting $n\to\infty $ we get $$\int_{h} ^{\infty} f(x) \, dx\leq h\sum_{r=1}^{\infty} f(rh) \leq \int_{0}^{\infty} f(x) \, dx$$ Letting $h\to 0^{+}$ we get the desired limit as $\int_{0}^{\infty} f(x) \, dx=\log 2$.

The above technique has also been used in this answer.

  • $\begingroup$ Hi Paramanand! (+1) … I just posted a solution that uses an alternative way forward. I hope you like it. ;-) $\endgroup$
    – Mark Viola
    Commented May 14, 2020 at 15:18

Note that we have

$$\begin{align} (1-t)\sum_{n=1}^\infty \frac{t^n}{1+t^n}&=(1-t)\sum_{n=1}^\infty \sum_{m=0}^\infty (-1)^mt^{n+nm}\tag1\\\\ &=(1-t)\sum_{m=0}^\infty (-1)^m \sum_{n=1}^\infty t^{(m+1)n}\tag2\\\\ &=(1-t)\sum_{m=0}^\infty (-1)^m \frac{t^{m+1}}{1-t^{m+1}}\\\\ &=\sum_{m=1}^\infty (-1)^{m-1}\frac{t^{m+1}}{\sum_{\ell=1}^mt^\ell} \end{align}$$

Now taking the limit as $t\to 1^-$ reveals

$$\begin{align} \lim_{t\to1^-}(1-t)\sum_{n=1}^\infty \frac{t^n}{1+t^n}&=\sum_{m=1}^\infty \frac{(-1)^{m-1}}m\tag3\\\\ &=\log(2) \end{align}$$

as expected!

NOTE $1$:

We first justify the interchange of series in going from $(1)$ to $(2)$. To do so, note that

$$\begin{align} \lim_{M\to\infty}\sum_{n=1}^\infty\sum_{m=0}^M (-1)^mt^{n+nm}&=\lim_{M\to\infty}\sum_{n=1}^\infty t^n \frac{1-t^{n(M+1)}}{1+t^n} \end{align}$$

For any fixed $t<1$, $\displaystyle \left|t^n \frac{1-t^{n(M+1)}}{1+t^n}\right|\le \frac{t^n}{1+t^n}$. Then, both the Dominated Convergence Test and the Weierstrass M-test, guarantee that

$$\begin{align} \lim_{M\to\infty}\sum_{n=1}^\infty\sum_{m=0}^M (-1)^mt^{n+nm}&=\lim_{M\to\infty}\sum_{n=1}^\infty t^n \frac{1-t^{n(M+1)}}{1+t^n}\\\\ &=\sum_{n=1}^\infty \frac{t^n}{1+t^n}\\\\ &=\sum_{n=1}^\infty\sum_{m=0}^\infty (-1)^mt^{n+nm}\tag4 \end{align}$$

Finally, we have

$$\begin{align} \lim_{M\to\infty}\sum_{n=1}^\infty\sum_{m=0}^M (-1)^mt^{n+nm}&=\lim_{M\to\infty}\sum_{m=0}^M \sum_{n=1}^\infty(-1)^mt^{n+nm}\\\\ &=\sum_{m=0}^\infty \sum_{n=1}^\infty(-1)^mt^{n+nm}\tag5 \end{align}$$

Noting that the right-hand sides are equal shows that the interchange of series is legitimate.

NOTE $2$:

To justify the interchange of the limit and the series on the left-hand side of $(3)$, simply note that for $t\le 1$


Hence, $\displaystyle \frac{t^{m+1}}{\sum_{\ell=1}^mt^\ell}\to 0$ uniformly as $m\to \infty$ (It is trivial to show that it is also monotonically decreasing). Dirichlet's Test guarantees then that

$$\lim_{t\to1^-}\sum_{m=1}^\infty (-1)^{m-1}\frac{t^{m+1}}{\sum_{\ell=1}^mt^\ell}=\sum_{m=1}^\infty (-1)^{n-1}\lim_{t\to1^-}\left(\frac{t^{m+1}}{\sum_{\ell=1}^mt^\ell}\right)$$

And we are done!

  • 1
    $\begingroup$ Cool! Very ingenious approach. +1 $\endgroup$
    – Paramanand Singh
    Commented May 14, 2020 at 16:22
  • $\begingroup$ @ParamanandSingh Thank you my friend! $\endgroup$
    – Mark Viola
    Commented May 14, 2020 at 16:27
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    $\begingroup$ +1. Very nice, can I ask you to justify some step, please? We know that the geometric series has that sum for $|t|<1$, are we supposing $t<1$ because we're interested at $t \to 1^{-}$ so it is allowed to suppose $t<1$? Why can we interchange the series in the second line (maybe for uniform convergence of geometric series)? In the third line it seems you've used again the sum of the geometric series, but $n$ starts from $1$ so shouldn't we have $\frac{1}{1-t^{m+1}}-1$? Why can we take the limit inside the series in the last step (maybe again uniforme convergence)? Thanks! $\endgroup$
    – ZaWarudo
    Commented May 21, 2020 at 20:31
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    $\begingroup$ @ZaWarudo Thank you and good catch on the typo of omission of the term $t^{m+1}$. I've edited edited accordingly. I've also added notes to explain justification for interchanging operations. $\endgroup$
    – Mark Viola
    Commented May 22, 2020 at 3:53

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