Prove that $\sum_{n=1}^\infty \frac{1}{n(n-x)}$ converges absolutely and uniformly on each [a,b]\N Let $a<b$, $a,b\in R$ Prove that the series $\sum_{n=1}^\infty \frac{1}{n(n-x)}$ converges absolutely and uniformly on [a,b]\N 
What is [a,b]\N? I am really confused. For absolute convergence I know we take the absolute of the series and for uniform we need Weistress or Abel's uniform convergence test.
Any help much appreciated!

Fact: Suppose $f_n$ is a sequence of real-valued functions defined on a set $E.$ Let $N\in \mathbb N.$ If $\sum_{n=N}^{\infty} f_n$ converges uniformly on $E,$ then $\sum_{n=1}^{\infty} f_n$ converges uniformly on $E.$ I'll use this below.
In our problem, it's enough to show that the series converges uniformly on $[-N,N]\setminus \mathbb N$ for each $N \in \mathbb N.$ Fix such an $N.$ Then for $n>N,$ we have
$$\left |\frac{1}{n(n-x)} \right| \le \frac{1}{n(n-N)}$$
for all $x \in [-N,N].$ Because $\sum_{n=N+1}^{\infty} 1/n(n-N) < \infty,$ we see $ \sum_{n=N+1}^{\infty} 1/n(n-x)$ converges uniformly on $[-N,N]$ by the Weierstrass M-test. This implies $ \sum_{n=1}^{\infty} 1/n(n-x)$ converges uniformly on $[-N,N]\setminus \mathbb N$ as desired.
