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?
Best Answer
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 $g_n(x)=x^{2n+1}$.