Suppose $$f(x)=\sum_{n=1}^\infty (-1)^{n+1} \ln (1+\frac{x}{n}), \quad x\in[0,\infty).$$ I need to show that $f$ is differentiable on $(0,\infty).$

proof: I try to show differentiability using the classical defintion of limit.

$$\lim _{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}$$ $$=\lim _{h\rightarrow 0} \bigg(\frac{\sum_{n=1}^\infty (-1)^{n+1} \ln (1+\frac{x+h}{n})-\sum_{n=1}^\infty (-1)^{n+1} \ln (1+\frac{x}{n})}{h}\bigg)$$ $$=\lim _{h\rightarrow 0} \sum_{n=1}^\infty (-1)^{n+1} \frac{\bigg(\ln (1+\frac{x+h}{n}) -\ln (1+\frac{x}{n})\bigg)}{h}$$ $$=\sum_{n=1}^\infty (-1)^{n+1} \lim_{h\rightarrow 0} \bigg(\frac{ \ln (\frac{n+x+h}{n}) -\ln (\frac{n+x}{n})}{h}\bigg)$$ $$=\sum_{n=1}^\infty (-1)^{n+1} \lim_{h\rightarrow 0} \bigg(\frac{1}{\frac{n+x+h}{n}.\frac{1}{n}}-\frac{1}{\frac{n+x}{n}.\frac{1}{n}}\bigg)\mbox{, By L'Hospital's rule.}$$ $$=\sum_{n=1}^\infty (-1)^{n+1} .0 =0$$

Thus, limit exists. Since $x$ was arbitrary, the claim follows.

Only problem I have is in justifying the exchange of limit and sum.

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When you used L'Hopital's rule, you should have differentiated with respect to $h$ on the top. As to justifying switching the limit and the sum, I haven't figured it out. The series converges conditionally, and they can be quite hard. – Stephen Montgomery-Smith Dec 20 '13 at 4:23
ya, my differentiation was stupid. I am realizing it now. – math Dec 20 '13 at 4:24
@mathandzen You're limit should not equal $0$ anyways because the limit should give you the expression for the derivative $f'(x)$. – andraiamatrix Dec 20 '13 at 4:47

$$\lim_{h\to 0} \sum_{n=1}^\infty (-1)^{n+1}\frac{\left(\ln(\frac {n+x+h}n) - \ln(\frac {n+x}n)\right)}{h} \\ = \lim_{h\to 0} \sum_{n=1}^\infty (-1)^{n+1}\frac{\ln(\frac {n+x+h}{n+x})}{h} \\ = \lim_{h\to 0} \sum_{n=1}^\infty (-1)^{n+1}\frac{\ln(1+\frac {h}{n+x})}{h}$$ Now use the formula $\ln(1+a) = a + O(a^2)$ as $a\to 0$. You then will get a term where the $h$'s cancel, plus a term that involves an absolutely converging series times $h$.

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Group two consecutive terms,

\begin{align} f(x) &= \sum_{n=1}^\infty (-1)^{n+1}\log \left(1+\frac{x}{n}\right)\\ &= \sum_{k=1}^\infty \left(\log \left(1+\frac{x}{2k-1}\right) - \log \left(1 + \frac{x}{2k}\right)\right). \end{align}

Now you have an absolutely and locally uniformly convergent series, and the series of the differentiated terms

$$\frac{1}{2k-1+x} - \frac{1}{2k+x} = \frac{1}{(2k-1+x)(2k+x)}$$

also converges absolutely and locally uniformly.

By standard arguments, $f$ is differentiable, with derivative

$$f'(x) = \sum_{k=1}^\infty \left(\frac{1}{2k-1+x} - \frac{1}{2k+x}\right).$$

If you wish, you can now split the terms again and write

$$f'(x) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n+x},$$

which converges pointwise to $f'(x)$. However, the absolutely convergent form with grouped terms allows easier manipulation.

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