Show that $\sum _{n=1}^{\infty }\:\frac{\left(-1\right)^n}{\:x+n}$ converges uniformly Show that $\sum _{n=1}^{\infty }\:\frac{\left(-1\right)^n}{\:x+n}$ converges uniformly on $[0, A]$ for any $A>0$ and the resulting function is differentiable on $(0,\infty)$ .
I  proved the first part by the alternative series test. Is it correct?
Thanks!
 A: The alternating series test only shows that the series converges; it says nothing about uniform convergence.
For example, the series
$$
f(x)=
\sum_{n=0}^\infty (-1)^n \frac{x^n}{n!}
$$
converges (pointwise) for all $x$ but does not converge uniformly.  You can see this because if for each small $\epsilon>0$ there is assigned an $N(\epsilon)$, then by taking $x > 2N(\epsilon)$ you have constructed a point at which $|f_N(x)-f(x)|>2^N>\epsilon$.
For $$f_N(x) =    \sum_{n=0}^N (-1)^n \frac{1}{x+n}$$ and  $$f(x) =    \sum_{n=0}^\infty (-1)^n \frac{1}{x+n}$$ 
The series does converge uniformly: Choose $N(\epsilon) = \frac1\epsilon +1$.
Then (for $x\geq 0$, which is the specified interval) because this is an alternating series with decreasing absolute value of terms, $|f_N(x)-f(x)|$ is less than or equal to the absolute value of first omitted term, which is 
$$
\frac1{x+N(\epsilon)} = \frac1{x+\frac1\epsilon+1} = \frac{\epsilon}{1+\epsilon x + \epsilon} < \epsilon
$$
Note that for this series, you can make $x$ as large as you want without having to change $N(\epsilon)$ so the series converges uniformly.
A: When I see an alternating series,
I like to combine
pairs of terms
to make all the terms
(I hope) of the same sign.
Your case:
In case it doesn't converge,
take a finite number of terms;
if it looks OK,
let the number of terms grow.
$\begin{array}\\
f_{2m}(x)
&=\sum _{n=1}^{2m }\:\frac{\left(-1\right)^n}{\:x+n}\\
&=\sum _{n=1}^{m }\left(\frac{-1}{x+2n-1}+\frac{1}{x+2n}\right)\\
&=\sum _{n=1}^{m }\frac{(x+2n-1)-(x+2n)}{(x+2n-1)(x+2n)}\\
&=\sum _{n=1}^{m }\frac{-1}{(x+2n-1)(x+2n)}\\
&=-\sum _{n=1}^{m }\frac{1}{(x+2n-1)(x+2n)}\\
\end{array}
$
Since each term
is less
in absolute value than
$\frac1{2n(2n-1)}$,
this converges absolutely
and uniformly,
so we can let
$m \to \infty$.
Note that this is legal
because each term
$\frac1{x+n} \to 0$.
You can now differentiate
term by term
to get a sum for
the derivative.
