Show that $\frac{1}{(n+1)!}(1+\frac{1}{(n+1)}+\frac{1}{(n+1)^2}+\frac{1}{(n+1)^3}+\cdots)=\frac{1}{n!n}$
Show that $\frac{1}{(n+1)!}(1+\frac{1}{(n+1)}+\frac{1}{(n+1)^2}+\frac{1}{(n+1)^3}+\cdots)=\frac{1}{n!n}$, (here $n$ is a natural number)
Maybe easy, but I cannot see it.
Thanks in advance! Alexander
Hint:
$$\sum_{k=0}^\infty \left(\frac1{n+1}\right)^k=\frac1{1-\frac1{n+1}}\ldots$$
Since $$1+x+x^2+\cdots=\frac1{1-x}$$
Therefore,
$$1+\frac1{n+1}+\frac1{(n+1)^2}+\cdots=\frac1{1-\frac1{n+1}}=\frac1{\frac{n+1-1}{n+1}}=\frac{n+1}{n}$$
Now,
$$\frac1{(n+1)!}\cdot(1+\frac1{n+1}+\frac1{(n+1)^2}+\cdots)=\frac1{(n+1)!}\cdot\frac{n+1}{n}=\frac1{n!n}$$