Continuity and differentiability of $f(x) = \frac{x}{1+x} + \frac{x}{(x+1)(2x+1)} + \frac{x}{(2x+1)(3x+1)}+...$ Here's the given function:

$$f(x) = \dfrac{x}{1+x} + \dfrac{x}{(x+1)(2x+1)} + \dfrac{x}{(2x+1)(3x+1)}+...$$

My question regarding this function: is this function considered a continuous and a differenciable function? Is it not continuous or differenciable at any specific points?
According to a question about this function, $f(x)$ is non-continuous. They haven't mentioned anything about it being differenciable or not.

My working: 
Simplyfing the series on the right hand side of the equation is quite straightforward-
$$f(x) = \dfrac{x}{1+x} + \dfrac{x}{(x+1)(2x+1)} + \dfrac{x}{(2x+1)(3x+1)}+...$$
$$\implies f(x) = \dfrac{1+x-1}{1+x} + \dfrac{(2x+1)-(x+1)}{(x+1)(2x+1)} + \dfrac{(3x+1)-(2x+1)}{(2x+1)(3x+1)}+...$$
$$\implies f(x) = 1 -\dfrac{1}{1+x}+\dfrac{1}{1+x} -\dfrac{1}{2x+1} +\dfrac{1}{2x+1}-\dfrac{1}{3x+1}...$$

$$\implies f(x) =1 $$

Since $f(x)$ is a constant function, it appears to be both continuous and differenciable $\forall x \in \mathbb R$. This is clearly incorrect, what am I doing wrong?

EDIT:
Is there a way to graph a function like $f(x)$?

Note that all the terms are $0$ when $x=0$!. Your calculation is correct if $x \neq 0$ and $x \neq -\frac 1 n$ for any $n$. Continuity and differentabilty are subject to intepretation. One interpretation is that since $f$ is not defined in any open interval containing $0$ it is not continuous, hence not differntiable, at $0$. 
