Can you prove that $\frac{1}{n}+\frac{1}{n^2}+\frac{1}{n^3}.... = \frac{1}{n-1}$? I'm a student and, while playing with my calculator, find out that: 
$$\frac{1}{n}+\frac{1}{n^2}+\frac{1}{n^3}.... = \frac{1}{n-1}$$
is it a well defined equivalence and what is its name, is there a proof for that?
if we put it this way:
$$1+\frac{1}{n}+\frac{1}{n^2}+\frac{1}{n^3}.... = \frac{n}{n-1}$$
what do you call the last term (the sum), the complementary inverse or reciprocal of 1/n?

assume $n>1$. Then
$$\sum_{k=0}^{K} \left(\frac{1}{n}\right)^k=1+\frac{1}{n}+...+\frac{1}{n^{K}}=L$$
Notice however, that $\frac{1}{n}\cdot L=\frac{1}{n}+..+\frac{1}{n^{K+1}}$
And then, that when you subtract them, all but the first and last term cancel: $$L-\frac{1}{n}\cdot L=1-\frac{1}{n^{K+1}}\implies L(1-\frac{1}{n})=1-\frac{1}{n^{K+1}} \implies L=\frac{1-1/(n^{K+1})}{(n-1)/(n)}=\frac{n(1-1/(n^{K+1}))}{(n-1)}$$
Now notice that as $K \to \infty$, $\frac{1}{n^{K+1}} \to 0$, so the term cancels. Thus: $$\sum_{k=0}^{\infty} \left(\frac{1}{n}\right)^k=\frac{n}{(n-1)}$$
This is a familiar trick to derive the more general geometric series equation:
geometric series
