<s>[INST] 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
 [/INST] 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:
$$\sum_{n=1}^{\infty} \frac{1}{(x+n)(x+n-1)} = \frac{1}{x} \, . $$
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