Evaluate $\frac{0!}{4!}+\frac{1!}{5!}+\frac{2!}{6!}+\frac{3!}{7!}+\frac{4!}{8!}+\cdots$

$$\frac{0!}{4!}+\frac{1!}{5!}+\frac{2!}{6!}+\frac{3!}{7!}+\frac{4!}{8!}+\frac{5!}{9!}+\frac{6!}{10!}+\cdots$$

This goes up to infinity. Trying finite cases may help.

My Attempt:It seems that it is going to be $\frac{1}{18}$. My calculations show that its going near $\frac{1}{18}$.

Solution 1:

An alternative approach to Behrouz' fine one through Euler's beta function. We have: $$\begin{eqnarray*} \sum_{n\geq 0}\frac{n!}{(n+4)!}=\sum_{n\geq 0}\frac{\Gamma(n+1)}{\Gamma(n+5)}&=&\frac{1}{\Gamma(4)}\sum_{n\geq 0}B(4,n+1)\\&=&\frac{1}{6}\int_{0}^{1}\sum_{n\geq 0}x^{n}(1-x)^3\,dx\\&=&\frac{1}{6}\int_{0}^{1}(1-x)^2\,dx\\&=&\frac{1}{6}\int_{0}^{1}x^2\,dx = \frac{1}{6}\cdot\frac{1}{3}=\color{red}{\frac{1}{18}}.\end{eqnarray*}$$

With the same approach it is straightforward to check that: $$ \sum_{n\geq 0}\frac{n!}{(n+k)!} = \color{red}{\frac{1}{(k-1)\cdot(k-1)!}}$$ for any $k\in\mathbb{N}^+$.

The same can be achieved by recognizing in the LHS a multiple of a telescopic series.

Solution 2:

Write $$\begin{align} {\frac{n!}{(n+4)!}}&=\frac{1}{(n+1)(n+2)(n+3)(n+4)}=\\&=\frac{A}{(n+1)(n+2)(n+3)}-\frac{B}{(n+2)(n+3)(n+4)}\end{align}$$

which gives $(n+4)A-B(n+1)=1 \implies A=B=1/3$

Then we get a telescoping series:

$$\begin{align}\sum\limits_{n=0}^{\infty }{\frac{n!}{(n+4)!}}=\frac{1}{3}\sum\limits_{n=0}^{\infty }\left({\frac{1}{(n+1)(n+2)(n+3)}}-\frac{1}{(n+2)(n+3)(n+4)}\right)=\frac{1}{18}\end{align}$$

Solution 3:

Just a tip; For natural number $n$, $$\sum_{k=1}^n\frac{1}{k(k+1)\cdots(k+m)}=\frac{1}{m}\left(\frac{1}{1\cdot 2\cdot \cdots\cdot m} -\frac{1}{(n+1)\cdot (n+2)\cdot \cdots\cdot (n+m)}\right) $$

Now, the given infinite sum is equal to

$$\sum_{k=1}^n \frac{(k-1)!}{(k+3)!}=\sum_{k=1}^n \frac{1}{k(k+1)(k+2)(k+3)}$$

by plugging in $m=3$ to the tip, we would get

$$\sum_{k=1}^n \frac{1}{k(k+1)(k+2)(k+3)}=\frac{1}{3} \left( \frac{1}{1\cdot2\cdot3} -\frac{1}{(n+1)(n+2)(n+3)}\right)$$

Take limit, $n \to \infty$. $$\sum_{k=1}^\infty \frac{1}{k(k+1)(k+2)(k+3)}=\frac{1}{3} \left( \frac{1}{1\cdot2\cdot3} -\lim_{n \to \infty}\frac{1}{(n+1)(n+2)(n+3)}\right)$$ $$=\frac{1}{18}$$

How to decompose the fraction; $$\frac{1}{k(k+1)\cdots(k+m)}=\frac{1}{m}\left(\frac{1}{k\cdot (k+1)\cdot \cdots\cdot (k+m-1)} -\frac{1}{(k+1)\cdot (k+2)\cdot \cdots\cdot (k+m)}\right)$$

Solution 4:

$A=\frac{1}{4*6}+\frac{1}{10*12}+\frac{1}{18*20}+\frac{1}{28*30}+\frac{1}{40*42}+\frac{1}{54*56}...$

$2A=\frac{2}{4*6}+\frac{2}{10*12}+\frac{2}{18*20}+\frac{2}{28*30}+\frac{2}{40*42}+\frac{2}{54*56}...$

$2A=\frac{1}{4}-\frac{1}{6}+\frac{1}{10}-\frac{1}{12}+\frac{1}{18}-\frac{1}{20}+\frac{1}{28}-\frac{1}{30}+\frac{1}{40}-\frac{1}{42}+\frac{1}{54}-\frac{1}{56}...$

Now I write the Negative fractions like this:

$$-\frac{1}{6}=\frac{1}{3}-\frac{1}{2}$$

$$-\frac{1}{12}=\frac{1}{4}-\frac{1}{3}$$

$$-\frac{1}{20}=\frac{1}{5}-\frac{1}{4}$$

$$-\frac{1}{30}=\frac{1}{6}-\frac{1}{5}$$

$$-\frac{1}{42}=\frac{1}{7}-\frac{1}{6}$$

$$-\frac{1}{56}=\frac{1}{8}-\frac{1}{7}$$

And do this up to infinity.The sum of all negative fractions are:

$\frac{1}{3}-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+\frac{1}{5}-\frac{1}{5}+\frac{1}{6}-\frac{1}{6}+\frac{1}{7}-\frac{1}{7}+\frac{1}{8}-\frac{1}{8}+...=-\frac{1}{2}$

Now we have:

$2A=\frac{1}{4}-\frac{1}{2}+\frac{1}{10}+\frac{1}{18}+\frac{1}{28}+\frac{1}{40}+...$

$2A=-\frac{1}{4}+\frac{1}{10}+\frac{1}{18}+\frac{1}{28}+\frac{1}{40}+...$

$2A=-\frac{1}{1*4}+\frac{1}{2*5}+\frac{1}{3*6}+\frac{1}{4*7}+\frac{1}{8*5}+...$

$6A=-\frac{3}{1*4}+\frac{3}{2*5}+\frac{3}{3*6}+\frac{3}{4*7}+\frac{3}{5*8}+...$

$6A=-1+\frac{1}{4}+\frac{1}{2}-\frac{1}{5}+\frac{1}{3}-\frac{1}{6}+\frac{1}{4}-\frac{1}{7}+\frac{1}{5}-\frac{1}{8}+...$

$6A=-1+\frac{1}{4}+\frac{1}{4}+\frac{1}{2}+\frac{1}{3}$

$6A=\frac{1}{3}$

$A=\frac{1}{18}$