Show that $\sum_{n=1}^\infty nx^{n-1}$ converges uniformly on $[0,\frac{9}{10}]$

Question

Show that $\sum_{n=1}^\infty nx^{n-1}$ converges uniformly on $[0,\frac{9}{10}]$

First I think we need to show pointwise convergence:

$$\sum_{n=1}^\infty nx^{n-1} = \lim_{N\to\infty} \sum_{n=1}^N nx^{n-1}$$

Since it's a finite sum we can do the following:

$$= \lim_{N\to\infty} \sum_{n=1}^N (x^n)' = \lim_{N\to\infty} \left(\sum_{n=1}^N x^n\right)^{'} = \lim_{N\to\infty} \frac{x^N-1}{x-1} = \frac{1-1}{x-1} = 0$$

Answer 1

Since $|nx^{n-1}| \le n(9/10)^{n-1}$ and $\sum_{n = 1}^\infty n(9/10)^{n-1}$ converges (by the ratio test), by the Weierstrass $M$-test, the series $\sum_{n = 1}^\infty nx^{n-1}$ converges uniformly on $[0,9/10]$.

Answer 2

Using geometric series we know exactly what your series should be

$$ \sum_{n=1}^\infty n x^{n-1} = \frac{1}{(1-x)^2}$$

Uniform convergence asked about the remaining terms and you can try to give uniform estimate

$$ \sum_{n=N+1}^\infty n x^{n-1} = x^N \left( \frac{1}{(1-x)^2} + \frac{N}{1-x} \right) \leq (\tfrac{9}{10})^N \left( \frac{1}{(1-\frac{9}{10})^2} + \frac{N}{1-\frac{9}{10}} \right) \to 0 $$

for all $0 \leq x \leq \frac{9}{10}$.