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

Question

I am trying to solve the following exercise and I would like to have a hint about the continuity part:

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 (NOTE: completed the exercise):

f defined We can rewrite $f$ as $f(x)=\sum_{n=0}^{\infty} (-1)^n \frac{1}{x+n}$ which converges by the Alternating Series Test so the function is surely defined for all $x>0$.

f continuous Now, let $\varepsilon>0$: then if we take $N>\frac{1}{\varepsilon}$ we have, for all $n,m\geq N, n>m$ and $x>0$ $|\frac{(-1)^{m+1}}{x+(m+1)}+\frac{(-1)^{m+2}}{x+(m+2)}+\dots+\frac{(-1)^n}{x+n}|\leq\frac{1}{x+(m+1)}<\frac{1}{x+m}<\frac{1}{m}<\varepsilon$ (note that wheter $m$ is even or odd makes no difference since $|-x|=|x|$) so the series converges uniformly Cauchy Criterion for Uniform Convergence of Series and $f$ is thus continuous by Term-by-Term Continuity Theorem.

f differentiable $|f'_n(x)|=|\frac{(-1)^{n+1}}{(x+n)^2}|=\frac{1}{(x+n)^2}\leq\frac{1}{n^2}$ and $\sum_{n=0}^{\infty}\frac{1}{n^2}$ is convergent so the series of derivatives $\sum_{n=0}^{\infty}\frac{(-1)^{n+1}}{(x+n)^2}$ is uniformly convergent on $(0,\infty)$ by Weierstrass M-Test and in particular on any interval $[a,b]\subset (0,\infty)$; also, since for any $x_0\in [a,b]$, $\sum_{n=0}^{\infty}f_n(x_0)$ is convergent by the Alternating Series Test we can conclude, by the Term-by-Term Differentiability Theorem, that on any interval $[a,b]\subset (0,\infty)$, $\sum_{n=0}^{\infty}f_n(x)$ converges uniformly to a differentiable function $f(x)=\sum_{n=0}^{\infty}f_n(x)=\sum_{n=0}^{\infty} (-1)^n \frac{1}{x+n}$ such that $f'(x)=\sum_{n=0}^{\infty}f'_n(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n+1}}{(x+n)^2}$.

Thank you.

Answer 1

Note for $x>0$, $$ 0< \frac{1}{x+n}-\frac{1}{x+n+1}+\frac{1}{x+n+2}-... \, ... \, ... <\frac{1}{x+n}, $$ so $$ \Big|\frac{1}{x+n}-\frac{1}{x+n+1}+\frac{1}{x+n+2}-... \, ... \, ... \Big|<\frac{1}{x+n}\leq \frac 1n. $$ So this series converges uniformly for $0\leq x<\infty$, so the sum is continuous.

Answer 2

Because $$\left| \sum\limits_{k=1}^{n} (-1)^k \right |\leqslant 1$$ and functional sequence $\frac{1}{x+n}$ uniformly with respect to $x$ tends to zero, then according to Dirichlet test series is uniformly converged, which gives continuity for every $x \in (0, +\infty)$.

Then, as formal derivative $$\sum\limits_{k=1}^{n} \frac{(-1)^{k+1}}{(x+n)^2} $$ also is converged uniformly by $\left| \frac{(-1)^{k+1}}{(x+n)^2}\right | \leqslant \frac{1}{n^2}$, then we can state that $f$ have derivative for every $x \in (0, +\infty)$ and holds $$f'(x)=\sum\limits_{k=1}^{\infty} \frac{(-1)^{k+1}}{(x+n)^2}$$ (btw last gives continuity as well).