Prove or disprove that $\sum_{n=1}^\infty n^3x^n$ is uniformly convergent on $x\in(-1,1)$
I started with proving that $\sum_{n=1}^\infty n^3x^n$ is convergent.
By ratio test:
$$ \lim_{n\to\infty}|\frac{a_{n+1}}{a_n}|=|\lim_{n\to\infty}\frac{(n+1)^3x^{n+1}}{n^3x^n}|=|x|\cdot|\lim_{n\to\infty}\frac{(n+1)^3}{n^3}|=|x|<1$$
so we get $\displaystyle\sum_{n=1}^\infty n^3x^n$ converges.
Now, I tried proving that it is uniformly convergent, I wanted to use the M-test, but I couldn't find a suitable series for the test. If the interval was $[-t,t]$ such that $t\in\mathbb{R}$, then I could do: $\sum_{n=1}^\infty n^3x^n\leq\sum_{n=1}^\infty n^3t^n$, and we know that $\sum_{n=1}^\infty n^3t^n$ converges because of the ratio test above. But this is not the case. Please help!
Best Answer
This is false.
Let $S_n(x)=\displaystyle\sum_{k=1}^n k^3x^k$, next, we show $S_n(x)$ is NOT uniformly convergent on $(-1,1)$. Consider the remainder
$$R_n(x)=\sum_{k=n+1}^\infty k^3x^k$$
If $S_n(x)$ is uniformly convergent, it means
$\forall\epsilon>0, \exists N$, $\forall n>N,\forall x\in (0,1)\Rightarrow|R_n(x)|<\epsilon, $.
We want to prove it is NOT uniformly convergent, so we negate it,
$\exists\epsilon>0, \forall N, \exists n>N, \exists x\in (0,1)\Rightarrow|R_n(x)|\ge\epsilon$.
Choose $\epsilon=1, n=N+1, x=1-\frac{1}{N+2}$,
$$|R_n(x)|=|R_{N+1}(x)|=\left|\sum_{k=N+2}^\infty k^3\left(1-\frac{1}{N+2}\right)^k\right|\ge (N+2)^3\left(1-\frac{1}{N+2}\right)^{N+2}\to\infty>1$$