Find sum of infinite anharmonic(?) series I need help with this:
$$
\frac{1}{1\cdot2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4\cdot5}+\frac{1}{3\cdot4\cdot5\cdot6}\dots
$$
I don't know how to count sum of this series. It is similar to standard anharmonic series so it should have similar solution. However I can't work it out.
 A: Using Euler's Beta function:
$$\begin{eqnarray*}\sum_{k=1}^{+\infty}\frac{1}{k(k+1)(k+2)(k+3)}&=&\sum_{k\geq 1}\frac{\Gamma(k)}{\Gamma(k+4)}=\frac{1}{\Gamma(4)}\sum_{k\geq 1}B(k,4)\\&=&\frac{1}{6}\sum_{k\geq 1}\int_{0}^{1}(1-x)^3 x^{k-1}\,dx\\&=&\frac{1}{6}\int_{0}^{1}(1-x)^2\,dx\\&=&\frac{1}{6}\int_{0}^{1}x^2\,dx = \color{red}{\frac{1}{18}}.\end{eqnarray*}$$
Using creative (not so much) telescoping, for $a_k=\frac{1}{k(k+1)(k+2)}$ we have $a_{k}-a_{k+1}=\frac{3}{k(k+1)(k+2)(k+3)}$, so $$\sum_{k\geq 1}\frac{1}{k(k+1)(k+2)(k+3)}=\frac{a_1}{3}=\frac{1}{3\cdot 3!}=\frac{1}{18}.$$
A: $$I=\dfrac2{n(n+1)(n+2)(n+3)}$$
$$2I=\dfrac{(n+2)(n+1)-n(n+3)}{n(n+1)(n+2)(n+3)}=\dfrac1{n(n+3)}-\dfrac1{(n+1)(n+2)}$$
$$6I=\dfrac1n-\dfrac1{n+3}-3\left(\dfrac1{n+1}-\dfrac1{n+2}\right)$$
Can you see the telescoping nature?
A: What you're evaluating is the infinite series
$$\sum_{n = 1}^\infty \frac{1}{n(n+1)(n+2)(n+3)}.$$
Since
$$\frac{1}{n(n+3)} = \frac{1}{3n}-\frac{1}{3(n+3)},$$
then
$$\frac{1}{n(n+1)(n+2)(n+3)} = \frac{1}{3n(n+1)(n+2)}-\frac{1}{3(n+1)(n+2)(n+3)}.$$
So your series telescopes to 
$$\frac{1}{3(1)(2)(3)} = \frac{1}{18}.$$
A: Use the identity 
\begin{align}
&\quad\sum_{n=1}^{\infty}\dfrac{1}{n(n+1)(n+2)\dots(n+k)}\\
\equiv&\quad\sum_{n=1}^{\infty}\dfrac{\Gamma(n)}{\Gamma(n+k+1)}\\
\equiv&\quad\frac{\Gamma (k)}{\Gamma (k+1)^2}\\
\equiv&\quad\dfrac{1}{k k!}, k\in \mathbb{N}, k\geq 1\\
\end{align}
A: $$\frac{1}{1\cdot2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4\cdot5}+\dots+\frac{1}{\frac{(n+4)!}{n!}}+\dots=\lim_{m\to\infty}\sum_{n=0}^{m}\frac{n!}{(n+4)!}=$$
$$\lim_{m\to\infty}\frac{m^3+9m^2+26m+18}{18(m+2)(m+3)(m+4)}=\lim_{m\to\infty}\frac{m^3+9m^2+26m+18}{18m^3+162m^2+468m+432}=$$
$$\lim_{m\to\infty}\frac{1+\frac{9}{m}+\frac{26}{m^2}+\frac{18}{m^3}}{18+\frac{162}{m}+\frac{468}{m^2}+\frac{432}{m^3}}=\frac{1+0+0+0}{18+0+0+0}=\frac{1}{18}$$
