Akbar Witt

2022-02-22

Find:
$\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k\left(k+1\right)\left(k+2\right)\left(k+3\right)}$

meizhen85ulg

There is a general solution for this type of problem:
$\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k\left(k+1\right)\cdot \dots \cdot \left(k+N\right)}$
Observe that
$\sum _{k\ge 1}\frac{1}{k\left(k+1\right)\cdot \dots \cdot \left(k+N\right)}=\sum _{k\ge 1}\frac{\left(k-1\right)!}{\left(k+N\right)!}$
$=\frac{1}{N!}\sum _{k\ge 1}\frac{\mathrm{\Gamma }\left(k\right)\mathrm{\Gamma }\left(N+1\right)}{\mathrm{\Gamma }\left(k+N+1\right)}=\frac{1}{N!}\sum _{k\ge 1}B\left(k,N+1\right)$
Consequently,
$\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k\left(k+1\right)\cdot \dots \cdot \left(k+N\right)}=\frac{1}{N!}\sum _{k\ge 1}{\int }_{0}^{1}{x}^{k-1}{\left(1-x\right)}^{N}dx$
$=\frac{1}{N!}{\int }_{0}^{1}\left(\sum _{k\ge 1}{x}^{k-1}\right){\left(1-x\right)}^{N}dx$
$=\frac{1}{N!}{\int }_{0}^{1}{\left(1-x\right)}^{N-1}dx$
$=\frac{1}{N\cdot N!}$
Here we have $N=3$. Hence, the answer is $\frac{1}{3\cdot 3!}=\frac{1}{18}$

an2gi2m9gg

You can also go the hard way: write
$\frac{1}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{x+2}+\frac{D}{x+3}$
so
$A\left(x+1\right)\left(x+2\right)\left(x+3\right)+Bx\left(x+2\right)\left(x+3\right)+Cx\left(x+1\right)\left(x+3\right)+Dx\left(x+1\right)\left(x+2\right)=1$
Now,
1. for
2. for
3. for
4. for
Thus
$\frac{1}{k\left(k+1\right)\left(k+2\right)\left(k+3\right)}=\frac{1}{6}\left(\frac{1}{k}-\frac{3}{k+1}+\frac{3}{k+2}-\frac{1}{k+3}\right)$
Hence your summation is $\frac{1}{6}$ of
$\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k}-3\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k+1}+3\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k+2}-\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k+3}=\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k}-3\sum _{k=2}^{\mathrm{\infty }}\frac{1}{k}+3\sum _{k=3}^{\mathrm{\infty }}\frac{1}{k}-\sum _{k=4}^{\mathrm{\infty }}\frac{1}{k}$
$=\left(1+\frac{1}{2}+\frac{1}{3}\right)-3\left(\frac{1}{2}+\frac{1}{3}\right)+3\frac{1}{3}$
$=\frac{1}{3}$
Therefore the sum of your series is
$\frac{1}{6}\cdot \frac{1}{3}=\frac{1}{18}$

Do you have a similar question?