$\sum\limits_{n\geq 0}\frac{1}{(n+1)(n+2)(n+3)}$ without using telescoping sums I'm really new to complex analysis and would like to see how one would go about finding a solution to an infinite series that looks like this: 
$$\sum\limits_{n\geq 0}\frac{1}{(n+1)(n+2)(n+3)}$$
or 
$$\sum\limits_{n\geq 0}\frac{1}{(n+1)(n+2)(n+3)(n+4)}$$
I'm interested in how we evaluate the residues as well. I understand that here we have only simple poles. These specific examples I made look a little tedious, but they have pretty simple solutions according to Wolfram. 
 A: This doesn't use complex analysis but it is to me the easiest and most natural way to evaluate these sums, so I will write it here. Let $f(x)=1/(1-x)=\sum_{n\geq0}x^n$, so that $D^{-1}f(x)=\sum_{n\geq0}\frac1{n+1}x^{n+1}$, and similarly $D^{-k}f(x)=\sum_{n\geq0}\frac{1}{(n+1)(n+2)\dotsm(n+k)}x^{n+k}$ for any integer $k>0$. Once we have this expression we then simply evaluate at $x=1$ (or take the limit as $x\to1$) to obtain our original sum.
As an example, say we wanted to evaluate $\sum_{n\geq0}\frac1{(n+1)(n+2)(n+3)}.$ Noting that all the $D^{-k}f(x)$ are $0$ at $x=0$, we see that
$$\begin{split}D^{-3}f(x)&=D^{-2}(-\log(1-x))\\
&=D^{-1}((1-x)\log(1-x)+x)\\
&=-\frac12\left(x-1\right)^2\log(1-x)+\frac14x(3x-2).\end{split}$$
So the value of the sum is nothing but the limit of this expression as $x\to1$, which you can work out (using e.g. the series expansion of $\log(1-x)$) to be $\dfrac14$, which is the desired answer.
A: My solution only goes for a even number of factors.
You can consider system of nested squares $\Gamma_n$: with centers in $0$ and vertices $z=\pm\pi(n+\frac{1}{2})\pm i\pi(n+\frac{1}{2})$. For them $d_n=\min\limits_{z\in\Gamma_n}|z|=\pi(n+\frac{1}{2})$, length of $\Gamma_n$ is $S_n=8\pi(n+\frac{1}{2})$, and $\frac{S_n}{d_n}=8$. You can prove that on this system of squares the $\text{ctg} z$ is bounded by some constant $c$ independent of $n$.
Now consider the function $f(z)=\frac{\pi^2\text{ctg} \pi z}{(z+1)(z+2)(z+3)(z+4)}$ and by Cauchy's residue theorem $\displaystyle\int\limits_{\Gamma_n^+}f(z)dz=2\pi i(\text{res}_0f(z)+\text{res}_{-1}f(z)+\text{res}_{-2}f(z)+\text{res}_{-3}f(z)+\text{res}_{-4}f(z)+\sum\limits_{k=-n}^n\text{res}_{ k}f(z))$. In the last sum $k\ne0,-1,-2,-3,-4$.
We calculate the residues $\text{res}_0f(z)=\frac{\pi}{24}$, $\text{res}_{-1}f(z)=-\frac{11\pi}{36}$, $\text{res}_{-2}f(z)=\frac{\pi}{4}$, $\text{res}_{-3}f(z)=\frac{\pi}{4}$, $\text{res}_{-4}f(z)=-\frac{11\pi}{36}$, $\text{res}_{k}f(z)=\frac{\pi}{(k+1)(k+2)(k+3)(k+4)}$.
So $\displaystyle\int\limits_{\Gamma_n^+}f(z)dz=2\pi i\left(-\frac{5\pi}{72}+\sum\limits_{k=-n}^{-5}\frac{\pi}{(k+1)(k+2)(k+3)(k+4)}+\sum\limits_{k=1}^{n}\frac{\pi}{(k+1)(k+2)(k+3)(k+4)}\right)$.
Further, $\left|\displaystyle\int\limits_{\Gamma_n^+}f(z)dz\right|\leq\frac{c\pi^2S_n}{(d_n-1)(d_n-2)(d_n-3)}\to0$, so $\sum\limits_{k=1}^{\infty}\frac{1}{(k+1)(k+2)(k+3)(k+4)}+\sum\limits_{k=-\infty}^{-5}\frac{1}{(k+1)(k+2)(k+3)(k+4)}=\frac{5}{72}$. Finally, The second sum is reduced to the first.
A: These particular examples can all be tackled with an elegant telescoping series. Observe that:
$$\frac{p-1}{(n+1)(n+p)}=\frac1{n+1}-\frac1{n+p}$$
Divide both sides by $(n+2)\dots(n+p-1)$ and you get a telescoping series, with the general result of:
$$\sum_{n\ge0}\frac1{(n+1)\dots(n+p)}=\frac1{(p-1)(p-1)!}$$
where $p=3,4$ are the cases given.
