What will be the formula for following infinite series?
$$1 + \frac{1!}{x+1} + \frac{2!}{(x+1)(x+2)}+ \cdots$$
$$ x\ge2 $$
up to infinite
What pattern i got :
coefficient of $ \frac{1!}{x+1}$ $(1+2+3+4+5+6+\cdots)$
coefficient of $ \frac{2!}{(x+1)(x+2)}$ is $ -(2+6+12+20+\cdots)$
coefficient of $ \frac{3!}{(x+1)(x+2)(x+3)} $ is $(3+12+30+\cdots)$
I want a formula as I put value of $n$ and get the answer.
Notice that $x+1=\frac{\Gamma{(x+2)}}{\Gamma{(x+1)}}$ and in general $(x+1)...(x+k)=\frac{\Gamma{(x+k+1)}}{\Gamma{(x+1)}}$ so your sum can be rewritten as $$\sum^{\infty}_{k=0}\frac{\Gamma{(k+1)}\Gamma{(x+1)}}{\Gamma{(x+k+1)}}=\sum^{\infty}_{k=0}\frac{\Gamma{(k+1)}\Gamma{(x+1)}}{\Gamma{(x+k+2)}}(x+k+1)$$ By the definition of the Beta function $$B(x,y)=\frac{\Gamma{(x)}\Gamma{(y)}}{\Gamma{(x+y)}}$$ we could rewrite the last summation in terms of the Beta function as $$\sum^{\infty}_{k=0}(x+k+1)\cdot B(k+1,x+1)$$ But $$B(x,y)=\int^{1}_{0}u^{x-1}(1-u)^{y-1}\,du$$ Therefore \begin{align}&\sum^{\infty}_{k=0}(x+k+1)\cdot B(k+1,x+1)=\sum^{\infty}_{k=0}(x+k+1)\cdot \int^{1}_{0}u^{k}(1-u)^{x}\,du\\&=\sum^{\infty}_{k=0}((x+1)\int^{1}_{0}u^{k}(1-u)^{x}\,du+k\int^{1}_{0}u^{k}(1-u)^{x}\,du)\\ &=(x+1)\sum^{\infty}_{k=0}\int^{1}_{0}u^{k}(1-u)^{x}\,du+\sum^{\infty}_{k=0}k\int^{1}_{0}u^{k}(1-u)^{x}\,du\\ &=(x+1)\int^{1}_{0}(\sum^{\infty}_{k=0}u^{k})(1-u)^{x}\,du+\sum^{\infty}_{k=1}k\int^{1}_{0}u^{k}(1-u)^{x}\,du\\ &=(x+1)\int^{1}_{0}(\sum^{\infty}_{k=0}u^{k})(1-u)^{x}\,du+\int^{1}_{0}u(\sum^{\infty}_{k=1}ku^{k-1})(1-u)^{x}\,du\\ &=(x+1)\int^{1}_{0}\frac{1}{1-u}(1-u)^{x}\,du+\int^{1}_{0}u \frac{1}{(1-u)^2}(1-u)^{x}\,du\\ &=(x+1)\int^{1}_{0}(1-u)^{x-1}\,dx+\int^{1}_{0}u(1-u)^{x-2}\,du\\ &=(x+1)\cdot \frac{1}{x}+B(2,x-1)\end{align} The last result is valid for any $x>1$.