Uniform convergence of series $\sum (-1)^n f_n(x)$

Question

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?

Answer 1

Here is another approach. You can use Dirichlet's Test for Uniform Convergence. In your case take

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

Answer 2

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 $L^{\infty}(\mathbb{R}^+)$.