# Convergence of series of functions implies uniform convergence of series of functions

So I have, for $x\in(0,\infty)$,

$$\sum\limits_{n=0}^\infty \frac{x}{(1+nx)(1+(n+1)x)}$$ Which is equal to $$\sum\limits_{n=0}^\infty \frac{1}{(1+nx)}-\frac{1}{(1+(n+1)x)}$$ Hence the partial sum formula is, for $k\in\mathbb{N}$, $$S_k(x)=1-\frac{1}{1+(k+1)x}$$ Hence $\lim\limits_{k\to\infty} S_k(x)=1$. Is this sufficient to conclude that the series converges uniformly on this interval?

Hint Uniform convergence on $(0,\infty )$ would mean that we can make
$|S_k(x) -1| = \frac{1}{1+(k+1)x}$
as small as we want by choosing $k$ independently of $x.$
• But, since we cannot choose $k$ independent of $x$, it does not converge uniformly, correct? In fact, by choosing $x_k=1/(k+1)$, $S_k(x_k) \to 1/2$. So it does not converge uniformly. Is my reasoning correct? – Hossmeister Apr 26 '18 at 2:24
• @Hossmeister You are correct. The convergence is not uniform on $(0,\infty)$. It is uniform on $[\delta,\infty)$ for all $\delta>0$. – Mark Viola Apr 26 '18 at 2:26