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

Question

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

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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?

Answer 1

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.

Answer 2

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!

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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.

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NOTE $2$:

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

$$\left|\frac{t^{m+1}}{\sum_{\ell=1}^mt^\ell}\right|\le\frac1m$$

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!