For what values of $x$ does $\sum_{n=1}^{\infty}\frac{1}{(x+n)(x+n-1)}$ converge? Please help me to find the values of $x$ in which the series $$\sum_{n=1}^{\infty}\frac{1}{(x+n)(x+n-1)}$$ converges? I applied some tests for it but...:(
Thank you
 A: Hint:
$$\frac{1}{(x+n)(x+n-1)}=\frac{1}{x+n-1}-\frac{1}{x+n}$$
A: We can use partial fractions to decompose this fraction. We need to find $A$ and $B$ such that
$$\frac{A}{x+n} + \frac{B}{x+n-1} \equiv \frac{1}{(x+n)(x+n-1)} \, . $$
If we cross multiply, we get $A(x+n-1)+B(x+n)\equiv1$. When $x=-n$ we get $-A\equiv1$ and when $x=1-n$ we get $B \equiv 1$. It now follows that:
$$\frac{1}{(x+n)(x+n-1)} \equiv \frac{1}{x+n-1} - \frac{1}{x+n} \, . $$
Let's examine these terms as $n$ advances from $1$. Substituting $n=1,2,3,\ldots$ gives:
$$\left( \frac{1}{x} - \frac{1}{x+1} \right) + \left( \frac{1}{x+1} - \frac{1}{x+2} \right) + \left( \frac{1}{x+2} - \frac{1}{x+3} \right) + \cdots = $$
$$\frac{1}{x} - \frac{1}{x+1} + \frac{1}{x+1} - \frac{1}{x+2} + \frac{1}{x+2} - \frac{1}{x+3}  + \cdots $$
You should be able to see that each term is cancelled by the very next term. For this to be well-defined we need the denominators to be non-zero and for the terms to tend to zero. With that in mind, $x$ can be anything but zero or a negative integer. With that in mind we have:
$$\sum_{n=1}^{\infty} \frac{1}{(x+n)(x+n-1)} = \frac{1}{x} \, . $$
A: Besides to @N.S. nice hit.  Think about $S_n$ and find the values that the following limit exists: $$\lim_{n\to\infty}S_n=1/x$$
A: The approach given by N.S. and Babak here is elegant. Here's something a little more direct, which doesn't require you to be clever.
First, figure out which values of $x$ would give you a term with zero in the denominator. Those are obviously bad values for $x$.
Next, note that all but finitely many terms will be positive (in fact all terms will be positive if $x$ is positive), so if the series diverges, the sum must be $+\infty$.
Let's assume $x>1$. If $x<1$, we can let $y$ be $x$ plus some large integer (so that $y>1$) and write
$$\sum_{n=1}^\infty\frac{1}{(x+n)(x+n-1)}=\sum_{n=1}^\infty\frac{1}{(y+n)(y+n-1)}+\text{finitely many terms}$$
Now $$\sum_{n=1}^\infty\frac{1}{(x+n)(x+n-1)}<\sum_{n=1}^\infty\frac{1}{(x+n-1)(x+n-1)}=\sum_{n=1}^\infty\frac{1}{(x+n-1)^2}<\sum_{n=1}^\infty\frac{1}{n^2}$$
That last sum is a convergent $p-$series.
