Uniform convergence of the sum $\sum_{n=0}^\infty {(n+1)x^n}$, $|x| < 1$

Question

Originally I needed to prove the continuity of the sum, but $(n+1)x^n$ is continuous, so I only need to show the uniform convergence of the sum. I know that the sum is equal to $\dfrac {1}{(x-1)^2}$ But

1. I couldn't find a way to show the uniform convergence using the definition or Cauchy uniform convergence theorem.
2. A sequence is uniformly convergent if and only if $\lim_{n \rightarrow \infty} {\sup_{|x|<1}|f_n(x) - f(x)| = 0}.$ But also, a function series is uniformly convergent only if its member (in this case $(n+1)x^n$) uniformly covnerges to $0$. But isn't $\sup_{|x|<1}|(n+1)x^n| = n+1$?

Answer 1

The series does not converge uniformly on $(-1,1)$. Prove by contradiction. Suppose the contrary that the series converges uniformly on $(-1,1)$. Denote $S_{n}(x)=\sum_{k=0}^{n}(k+1)x^{k}$. For $\varepsilon=\frac{1}{2}$, there exists $N\in\mathbb{N}$ such that $\left|S_{n}(x)-S_{m}(x)\right|<\varepsilon$ whenever $m,n\geq N$ and $x\in(-1,1)$. In particular, $\left|(N+2)x^{N+1}\right|=\left|S_{N+1}(x)-S_{N}(x)\right|<\varepsilon$ for all $x\in(-1,1)$. Note that $x^{N+1}\rightarrow1$ as $x\rightarrow1-$. Therefore, by letting $x\rightarrow1-$, we obtain $(N+2)\leq\varepsilon$, which is a contradiction.

Answer 2

Take $x_0\in(-1,1)$. Then take $r>0$ such that $x_0\in(-r,r)$. Your series converges uniformly on $(-r,r)$, by the Weierstrass $M$-test. Therefore, it's sum is continuous at $x_0$.

But the convergence is not uniform on $(-1,1)$; a sequence of bounded functions cannot converge uniformly to an unbounded one.

And, in this context, asserting that the convergence is uniform means that$$(\forall\varepsilon>0)(\exists N\in\Bbb N)(\forall n\in\Bbb N)(\forall x\in(-r,r)):n\geqslant N\implies\left|\frac1{(x-1)^2}-\sum_{k=0}^n(k+1)x^k\right|<\varepsilon.$$

Answer 3

You're trying to prove something that's false. If you had instead $|x|<L<1$ for some $L>0$, then the sum is indeed uniformly convergiant via Weierstrass M test with $M_n:=(n+1)L^n$.

In your case, suppose the sum is indeed uniformly convergent. Fix $\epsilon=1$. Then there exists an $N>0$ such that $|\sum_{n=N+1}^\infty (n+1)x^n|<1$ for all $|x|<1$. This is clearly false if $x$ is close enough to 1, e.g. just consider real $x$ very close to 1 from the left.