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?
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.