$\sum_{n=1}^{\infty} \frac{x}{(1+(n-1)x)(1+nx)}$ continuity

Question

I wish to check for which $x$ the sum:

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

is continuous.

my attempt:

this is a telescopic sum:

$\sum_{n=1}^{\infty} \frac{x}{(1+(n-1)x)(1+nx)}=\sum_{n=1}^{\infty}\frac{1}{1+(n-1)x)}-\frac{1}{1+nx}=1-\lim_{n\rightarrow \infty}\frac{1}{1+nx}$

So we get that:

$f(x)=\begin{cases} 1 && x\ne 0 \\ 0 && x=0 \end{cases}$

Therefore, I think I can conclude the function is continuous for $x\ne 0$.

My book claims the sum is continuous for $x\gt 0$, why is it not continuous for $x\lt 0$?

Have I made any mistakes?

Answer 1

You are right: the sum is discontinuous at $0$ and only there. There is no reason to distinguish that case $x>0$ from the case $x<0$.

Answer 2

According to theorem 1 for f(x) to be continuous f(x)=$\sum_{n=1}^\infty f_n(x)$; $f_n(x) $ need to be continous. $$f(x)=\sum_{n=1}^{\infty} \frac{x}{(1+(n-1)x)(1+nx)}= \sum_{n=1}^\infty (\frac{1}{1+(n-1)x}-\frac{1}{1+nx})$$ f(x) is not defined for for $x = -1/n$ for $n\in N$

Theorem 1: Let $(f_n(x))_∞^n$ be a sequence of functions with common domain X and suppose that $f_n$ is continuous at $c∈X$ for all $n∈N$. If $∑_n^∞ f_n(x)$ uniformly converges to the sum function f(x) then f is continuous at c.