Uniform convergence of the following series: Does the following series converge uniformly ?$$\sum_{n=0}^{\infty} \frac{-1^n} {x+n} \ for \ (x\in R^+)$$
My thought:
for pointwise convergence: $$ \lim_{n \rightarrow \infty} \frac{-1^n} {x+n} =0 $$
so $$|f_n(x) - f(x)|=|\frac{-1^n} {x+n}|$$
now for the function $$g(x)=\frac{-1^n} {x+n} \implies g'(x)= \frac {-(-1)^n}{(x+n)^2}\neq0 $$
how can I proceed further ? 
what I am trying to do here is to find the critical point to maximize the function $g(x)$ but clearly there is no critical point as the derivative can't be zero.
 A: You are mistakenly trying to prove that the series converges uniformly by analyzing the convergence of the term sequence. For a series $\sum f_n(x)$ to converge uniformly it is necessary but not sufficient that $f_n(x) \to 0$ uniformly.
In this case, for all $x \in \mathbb{R}^+$ we have
$$|f_n(x)| =\left|\frac{(-1)^n}{x + n} \right|  \leqslant \frac{1}{n}$$
and the convergence $f_n(x) \to 0$ is uniform.  This only  means we can't yet rule out uniform convergence of the series.
The  series does, in fact, converge uniformly by Dirichlet's test since the partial sums $\sum_{n=0}^m (-1)^n$ are uniformly bounded and $1/(x+n)$ converges monotonically and uniformly to $0$ for $x \in \mathbb{R}^+$.
A: Note that
$$
\begin{align}
\sum_{k=2n}^\infty\frac{(-1)^k}{k+x}
&=\sum_{k=n}^\infty\left(\frac1{2k+x}-\frac1{2k+1+x}\right)\\
&=\sum_{k=n}^\infty\frac1{(2k+x)(2k+1+x)}\\
&\le\sum_{k=n}^\infty\frac1{(2k-1)(2k+1)}\\
&=\frac1{4n-2}
\end{align}
$$
and
$$
\begin{align}
\sum_{k=2n+1}^\infty\frac{(-1)^k}{k+x}
&=-\sum_{k=n}^\infty\left(\frac1{2k+1+x}-\frac1{2k+2+x}\right)\\
&=-\sum_{k=n}^\infty\frac1{(2k+1+x)(2k+2+x)}\\
&\ge-\sum_{k=n}^\infty\frac1{2k(2k+2)}\\
&=-\frac1{4n}
\end{align}
$$
Therefore,
$$
\left|\,\sum_{k=n}^\infty\frac{(-1)^k}{k+x}\,\right|\le\frac1{2n-2}
$$
independent of $x\ge0$.
