$\frac{1}{n+1}+\frac{1}{(n+1)^2}...=\frac{1}{n}$? On a problem book solution I was faced with the following step:

$$\frac{1}{n+1}+\frac{1}{(n+1)^2}...=\frac{1}{n}$$

I identified $\frac{1}{n+1}+\frac{1}{(n+1)^2}...$ as a geometric series so the sum would be $\frac{1}{1-r}$ so that $\frac{1}{1-\frac{1}{n+1}}=1+\frac{1}{n}$. I do not understand what I am doing worng.
Question:
What am I doing wrong?
Thanks in advance!

If you know that $a+ar+ar^2+ar^3+\cdots = \dfrac{a}{1-r}$ for $|r|\lt 1$
then here you have $a= \dfrac1{n+1}$ and $r=  \dfrac1{n+1}$  
so the sum is $\dfrac{\frac1{n+1}}{{1-\frac1{n+1}}}=\dfrac{1}{n+1-1}=\dfrac{1}{n}$
