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

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.

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.
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).