Prove $g(x) = \sum_{n=1}^{\infty} \frac{d}{dx}(-1)^n(4x)^n$ is uniformly continuous on $[-a,a]$ for each $a$ where $0 \lt a \lt \frac{1}{4}$.

Question

Problem: Prove $g(x) = \sum_{n=1}^{\infty} \frac{d}{dx}(-1)^n(4x)^n$ is uniformly continuous on $[-a,a]$ for each $a$ where $0 \lt a \lt \frac{1}{4}$.

I tried to use thee following theorem which states that

if $g$ is convergent at $x=x_0 \ne 0$, then it is uniformly convergent on $[-x_1, x_1], 0 \lt x_1 \lt |x_0|$ .

Now $g(x) = \sum_{n=1}^{\infty} \frac{d}{dx}(-1)^n(4x)^n = \sum_{n=1}^{\infty} 4n(-1)^n(4x)^{n-1}$.

But for $x=\frac{1}{4}$, $g\left(\frac{1}{4} \right) = \sum_{n=1}^{\infty} 4n(-1)^n$ diverges. Also for $x = \frac{-1}{4}$.

So it looks like I can't use the theorem stated above.

The other approach I tried is the Weierstrass M test. Here is my attempt:

Define $M_n = n(4x)^n$. Then each term of the series is less than $M_k$. But the $M$ series itself doesn't seem to converge. That is, for $0 \lt x \lt \frac{1}{4}$, $4x \lt 1$, so $\sum_{n=1}^{\infty} M_n = \sum_{n=1}^{\infty} n(4x)^n$ = convergence?

This is where I am stuck. Is it possible to show the $M$ series is convergent? If not, can I come up with another definition of $M$ which would work? Or any other approach?

Thanks

Answer 1

The $g(x)=\sum_{n=1}^{\infty}(-1)^nn(4x)^{n-1}$ is bounded by $\sum_{n=1}^{\infty}n(4a)^{n-1}<\infty$ and continuous on $[-a,a]$ as a uniform limit of continuous functions(the partial sums) thus uniformly continuous.