Proof of the differentiability $\sum\limits_{n=1}^\infty (-1)^{n+1}\ln(1+\frac{x}{n}) $

Question

I have a problem to prove the differentiability of:

$$\ \sum_{n=1}^\infty (-1)^{n+1}\ln(1+\frac{x}{n}) ~~~~x\in[0, \infty)$$

I know that I have to prove the convergency, but I don't know how to deal with it. I was thinking of:

$$\ \sum_{n=0}^\infty (-1)^{n}\ln(1+\frac{x}{n-1}) $$

In order to use Leibnitz criterion however I don't think that it would work. Maybe I have to separate that to sequences for even and odd n? Additionaly, $\ (-1)^n $ confuses me as it comes to differentiation. I would appreciate your suggestions.

Answer 1

Hints: Consider $\sum (-1)^{n+1} [\ln (1+\frac x n) -\frac x n]$. Use the fact that $\frac {\ln (1+t)-t} {t^{2}} \to -\frac 1 2$ as $t \to 0$ to show that this series converges uniformy on $[0,A]$ for any $A$. Hence the original series converges uniformly iff $\sum (-1)^{n+1} \frac x n$ does. Since $\sum (-1)^{n+1} \frac 1 n$ converges we have proved that the given series is uniformly convergent on $[0,A]$. Next consider the differentiated series $\sum (-1)^{n+1} \frac 1 {n+x}$. Once again add and subtract $\sum (-1)^{n+1} \frac 1 {n}$ to show that the differentiated series also converges uniformly on $[0,A]$ for any $A$. This implies that the sum of the original series is differentiable.

I am using the following elementary fact:

If $s_n(x) \to f(x)$ uniformly on compact sets and If $s_n'(x) \to g(x)$ uniformly on compact sets then $f$ is differentiable and $f'(x)=g(x)$.

Answer 2

Since $|\ln(1+x/n)|$ is strictly decreasing to $0$ as $n\to\infty$, then

$$f(x):=\sum_{n=1}^\infty(-1)^{n+1}\ln\left(1+\frac xn\right)$$

converges by the alternating series test. By the limit definition of the derivative, we then have

\begin{align}f'(x)&=\lim_{h\to0}\frac{f(x+h)-f(x)}h\\&=\lim_{h\to0}\frac1h\sum_{n=1}^\infty(-1)^{n+1}\left[\ln\left(1+\frac{x+h}n\right)-\ln\left(1+\frac xn\right)\right]\tag1\\&=\lim_{h\to0}\frac1h\sum_{n=1}^\infty(-1)^{n+1}\ln\left(1+\frac h{x+n}\right)\tag2\\&=\lim_{h\to0}\frac1h\sum_{n=1}^\infty(-1)^{n+1}\left[\frac h{x+n}+g\left(\frac h{x+n}\right)\right]\tag3\\&=\lim_{h\to0}\sum_{n=1}^\infty(-1)^{n+1}\left[\frac1{x+n}+\frac1hg\left(\frac h{x+n}\right)\right]\tag4\\&=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{x+n}\tag5\end{align}

which converges again by the alternating series test, where

$(1):$ Substituting in the definition of $f$ and subtracting termwise.

$(2):$ Applying log rules and simplifying to get $$\ln\left(1+\frac{x+h}n\right)-\ln\left(1+\frac xn\right)=\ln\left(\frac{1+\frac{x+h}n}{1+\frac xn}\right)=\ln\left(\frac{x+n+h}{x+n}\right)=\ln\left(1+\frac h{x+n}\right)$$

$(3):$ Taylor expand the logarithm with the remainder term: $$\ln(1+t)=t+g(t)$$ $$|g(t)|\le\frac{t^2}{2(1+t_0)^2}\tag{$-1<t_0\le t$}$$ where $t=h/(x+n)$.

$(4):$ Divide the terms by $h$.

$(5):$ Take the limit to get \begin{align}\lim_{h\to0}\left|\sum_{n=1}^\infty\frac{(-1)^{n+1}}hg\left(\frac h{x+n}\right)\right|&\le\lim_{h\to0}\sum_{n=1}^\infty\left|\frac{(-1)^{n+1}}hg\left(\frac h{x+n}\right)\right|\\&\le\lim_{h\to0}\sum_{n=1}^\infty\frac{|h|}{2(1+t_0)(x+n)^2}\\&=\lim_{h\to0}|h|S(x)\\&=0\end{align} where $S(x)=\sum_{n=1}^\infty\frac1{2(1+t_0)^2(x+n)^2}$ converges by comparison to $\sum_{n=1}^\infty\frac1{n^2}$.