# How come $\left(\frac{n+1}{n-1}\right)^n = \left(1+\frac{2}{n-1}\right)^n$?

I'm looking at one of my professor's calculus slides and in one of his proofs he uses the identity:

$\left(\frac{n+1}{n-1}\right)^n = \left(1+\frac{2}{n-1}\right)^n$

Except I don't see why that's the case. I tried different algebraic tricks and couldn't get it to that form.

What am I missing?

Thanks.

Edit: Thanks to everyone who answered. Is there an "I feel stupid" badge? I really should have seen this a mile a way.

• Because $n+1=(n-1)+2$. – vadim123 Feb 23 '15 at 14:39

Just write $$\left(\frac{n+1}{n-1}\right)^n = \left(\frac{n-1+2}{n-1}\right)^n =\left(1+\frac{2}{n-1}\right)^n$$
$1+\frac{2}{n-1}=\frac{n-1}{n-1}+\frac{2}{n-1}=\frac{n+1}{n-1}$
Note that $$\frac{n+1}{n-1} = \frac{n-1+2}{n-1} = \frac{n-1}{n-1} + \frac{2}{n-1} = 1 + \frac{2}{n-1}.$$
$(1 + \frac {2}{n-1})^n = (\frac {n-1 +2}{n-1})^n = (\frac{n+1}{n-1} )^n$