Uniform convergence of series $\sum (-1)^n f_n(x)$ Check the uniform convergence of the series $$\sum (-1)^n  f_n(x)$$
where $$f_n(x) = \ln \left( 1+ \frac{x}{n(1+x)} \right)$$
Do i need to sum them two-by-two and obtain something neat?
Also why does the last inequality here
https://math.stackexchange.com/a/130031/45470
hold? If so, can i make use of it?
 A: By summing them two-by-two we have:
$$\begin{eqnarray*}\sum_{n\geq 1}(-1)^n f_n(x) &=& \sum_{m\geq 1}\log\left(\frac{2m+(2m+1)x}{2m+2mx}\cdot\frac{(2m-1)+(2m-1)x}{(2m-1)+2mx}\right)\\&=&\sum_{m\geq 1}\log\left(\frac{1+\left(1+\frac{1}{2m}\right)x}{1+\left(1+\frac{1}{2m-1}\right)x}\right)\end{eqnarray*}$$
and if we set:
$$ g_m(x) = \log\left(\frac{1+\left(1+\frac{1}{2m}\right)x}{1+\left(1+\frac{1}{2m-1}\right)x}\right) $$
we have that over $\mathbb{R}^+$ we have to deal with a decreasing function between:
$$ \log\left(1-\frac{1}{4m^2}\right)\leq g_m(x)\leq 0. $$
Since by the Wallis product:
$$\prod_{m=1}\left(1-\frac{1}{4m^2}\right) = \frac{2}{\pi},$$
we have:
$$\log 2-\log\pi \leq \sum_{n\geq 1}(-1)^n f_n(x) \leq 0 $$
regardless of $x\in\mathbb{R}^+$. Summing two-by-two is legit since for any $x\in\mathbb{R}^+$ we have:
$$ 0\leq \log\left(1+\frac{x}{n(1+x)}\right)\leq\log\left(1+\frac{1}{n}\right)\leq\frac{1}{n},$$
hence $\left\{\sum_{n=1}^{N}(-1)^n f_n(x)\right\}_{N\in\mathbb{N}}$ is a Cauchy sequence in $L^{\infty}(\mathbb{R}^+)$.
A: Here is another approach. You can use Dirichlet's Test for Uniform Convergence. In your case take 

$$a_n(x)=(-1)^n\,\quad  b_n(x) = \ln\left( 1+\frac{x}{n(1+x)}\right). $$

