<s>[INST] $\frac{1}{n+1}+\frac{1}{(n+1)^2}...=\frac{1}{n}$? 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!
 [/INST] How I recommend approaching these problems:
Use the fact that if $|x| < 1$, 
$$1 + x + x^2 + \cdots = \dfrac{1}{1-x}\text{.} \tag{*}$$
Then, factor to rewrite the problem so that it is in terms of the equation (*).

Observe that if I factor out $\dfrac{1}{n+1}$ that 
$$\dfrac{1}{n+1} + \dfrac{1}{(n+1)^2} + \cdots = \dfrac{1}{n+1}\left(1+\dfrac{1}{n+1}+\dfrac{1}{(n+1)^2}+\cdots \right)\text{.}\tag{A}$$
The sum
$$1+\dfrac{1}{n+1}+\dfrac{1}{(n+1)^2} + \cdots = \dfrac{1}{1-\frac{1}{n+1}} = \dfrac{1}{(n+1-1)/(n+1)} = \dfrac{n+1}{n}\tag{B}$$
as long as $$\left|\dfrac{1}{n+1} \right| < 1 \implies |n+1| > 1 \implies n+1 > 1 \text{ and } -(n+1) < -1 \implies n > 0\text{.}$$
Thus, as long as $n > 0$, combine (A) and (B) to get
$$\dfrac{1}{n+1} \cdot \dfrac{n+1}{n} = \dfrac{1}{n}$$
as desired. 
Please be sure to note that this is only true when $n > 0$. I have seen mistakes in the past when people ignore assumptions made when summing infinite geometric series.
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