Testing for Uniform Convergence of the sum of an Alternating Series. I'm still trying to get used in understanding the concept behind uniform convergence, so there's another questions which I'm currently have trouble trying to answer. 
Suppose there's a series $$\sum_{k=0}^{\infty}(-1)^k\frac{x^{2k+1}}{2k+1}$$ and x is such that $-1 \leq x \leq 1$.
My first attempt was to use the Weierstrass' M Test but I can only seem to find $M_k$ such that $$M_k=\frac{1}{2k+1}$$. However, after a comparison test $\sum_{k=0}^{\infty}M_k$ doesn't converge.
I tried to find a partial sum of $\sum_{k=0}^{\infty}(-1)^k\frac{x^{2k+1}}{2k+1}$ to work with similar to the last question I posted such as $$S_n=\sum_{k=0}^{n}(-1)^k\frac{x^{2k+1}}{2k+1}$$ where I realise the last term could actually be an even number n=2z or an odd number n=2z+1 and as a result could have an impact on the sign of the last term.
My thinking was to derive a Sum such that $$S_{2n+1}=\sum_{k=2n}^{2n+1}(-1)^k \frac{x^{2k+1}}{2k+1}=-\frac{x^{4n+3}}{4n+3}$$ and attempt prove uniform convergence of that.
Would this be an appropriate method or am I going the wrong way about this completely?
 A: You can prove the convergence using the following Abel's test.
Abel's test: Let the series of functions $\sum f_n$ be uniformly convergent on $[a,b]$. Let the sequence of functions $\{g_n\}$ be monotonic for every $x\in [a,b]$ and uniformly bounded on $[a,b]$. Then the series $\sum f_ng_n$ is uniformly convergent.
For your problem, take $f_n=\frac{(-1)^n}{2n+1}$ and $g_n(x)=x^{2n+1}$.
A: This problem isn't quite difficult. First, the series is a power series, with radius of convergence $R=1$, to obtain this you can use Cauchy-Hadamard formula or Ratio Test.Finally you use Abel's Theorem: if $f(x)=\sum_{n=0}^\infty a_n x^n$ ($a_n,x\in\mathbb{R}$) has convergence ratio $R$, and the numerical serie $\sum_{n=0}^\infty a_n R^n$ converges ($\sum_{n=0}^\infty a_n (-R)^n$) then $\sum_{n=0}^\infty a_n x^n$ is uniform convergent in $[0,R]$ ($[-R,0]$).
With this at hand: evaluating in $x=1$ we have $\displaystyle\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}$ and this series is convergent -Leibniz Criteria-. In the other case, $x=-1$ we get $\displaystyle\sum_{n=0}^\infty(-1)^n\frac{(-1)^{2n+1}}{2n+1}=\sum_{n=0}^\infty(-1)^{n-1}\frac{1}{2n+1}$ and this series is convergent, again by Leibniz Criteria.
I hope it will be ok for you. 
