Show if sum of functions is defined Show $$\sum_{n=0}^{\infty} \frac{(-1)^n}{x+n}$$ is defined for all $x > 0$ and decide if it is continuous and if it is differentiable on $(0, \infty)$.  My thought is to use the Weirstrass M test. The only problem is that the bound depends on $x$, and I don't know if the theorem works in that case. Any ideas?
 A: The function $f(x)$ as represented by the series
$$f(x)=\sum_{n=0}^\infty \frac{(-1)^n}{x+n}$$
exists for all $x\ne 0$.  To show this, we can apply Leibniz's Test or alternatively Dirichlet's Test. 

CONTINUITY FOR $\displaystyle x\in (0,\infty)$:
To show that $f(x)$ is continuous on $x\in (0,\infty)$, we write for $x+h>0$
$$\begin{align}
|f(x+h)-f(x)|&=\left|\sum_{n=0}^\infty\frac{(-1)^{n+1}h}{(x+h+n)(x+n)}\right|\\\\
&\le |h|\sum_{n=0}^\infty \frac{1}{(x+h+n)(x+n)}\\\\
&\le |h|\left(\frac{1}{x(x+h)}+\sum_{n=1}^\infty \frac{1}{n^2}\right)\\\\
&\to 0\,\,\text{as}\,\,h\to 0
\end{align}$$

DIFFERENTIABLILITY FOR $\displaystyle x\in (0,\infty)$:
To show that $f'(x)$ exists and is given by $f'(x)=\sum_{n=0}^\infty \frac{(-1)^{n+1}}{(x+n)^2}$ for $x\in (0,\infty)$, we can write for $x+h>0$
$$\begin{align}
\left|\frac{f(x+h)-f(x)}{h}-\sum_{n=0}^\infty \frac{(-1)^{n+1}}{(x+n)^2}\right|&=\left|\sum_{n=0}^\infty (-1)^{n+1}\left(\frac{1}{(x+h+n)(x+n)}-\frac{1}{(x+n)^2}\right)\right|\\\\
&\le |h|\sum_{n=0}\frac{1}{(x+h+n)(x+n)^2}\\\\
&=|h|\left(\frac{1}{x^2(x+h)}+\sum_{n=1}^\infty \frac{1}{(x+h+n)(x+n)^2}\right)\\\\
&\to 0\,\,\text{as}\,\,h\to 0
\end{align}$$
And we are done!
A: Just for my own enjoyment, the series can be evaluated as follows:
$$\left|\sum_{n=0}^N(-1)^n\right|<2,\ \lim_{n\to\infty}\frac1{x+n}=0,\ \frac1{x+n}\text{ is monotonically decreasing after $n>-x$}$$
Thus, it converges forall $x\notin\{0,-1,-2,-3,\dots\}$ by the Dirichlet test.
It is easy enough to see that
$$\frac1{1-r}=\sum_{n=0}^\infty r^n \forall\ |r|<1$$
$$\frac{r^{x-1}}{1+r}=\sum_{n=0}^\infty(-1)^nr^{n+x-1}$$
and with some rigor,
$$\begin{align}\int_0^1\frac{r^{x-1}}{1+r}dr&=\int_0^1\sum_{n=0}^\infty(-1)^nr^{n+x-1}dr\\&=\sum_{n=0}^\infty\int_0^1(-1)^nr^{n+x-1}dr\\&=\sum_{n=0}^\infty{(-1)^n\over x+n}\end{align}$$
So for the intended purposes, it would be useful to use

$$\sum_{n=0}^\infty\frac{(-1)^n}{x+n}=\int_0^1\frac{r^{x-1}}{1+r}dr$$

which converges for $x>0$.  Adjusting for $x<0$ can be done simply by reindexing.
