I am working on exercise 6.4.6. from Abbott's Understanding Analysis 2nd ed. The statement is the following:
Let $f(x)=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+4}-\dots.$
Show $f$ is defined for all $x>0$. Is $f$ continuous on $(0,\infty)$? How about differentiable?
My solution:
We know that the alternating $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}$ converges. We can also write the sum of the statement as $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{x+n-1}$, and use $f_{n}(x)$ for a specific term of the series.
a) If $x>1$, the series $f$ differs by a positive number in the denominator, so every series term is smaller than every term of the alternating series and hence would converge.
If $x\in(0,1)$, then after $n=2$ we have the same as above, and for $n=1$, we are just adding a constant term to the series, giving us a convergent series.
Combining these results, we have that the series of the statement converges uniformly.
b) Combining uniform convergence and the fact that for a specific $n$, $f_{n}(x)$ is continuous, we have that by the term-by-term continuity theorem, $f(x)$ is continuous.
c) If we take the derivative of $f_{n}(x)$ for a specific $n$ we get: $f_{n}'(x)=\frac{(-1)^{n}}{(x+n-1)^{2}}$. Performing a similar analysis as in a), we have that the series of derivatives converges uniformly, and conclude that $f(x)$ is differentiable by the term-by-term differentiatiability theorem.
I would be grateful if you check whether my solution is correct. In particular, I am a bit doubtful of the conclusion in italics in part a). The problem I see is that I cannot use Weierstrass M-test directly because of the alternating nature of the series I am comparing with.