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Is this a new mathematical concept?

$$ \frac{1}{n} + \frac{1}{n^2} + \frac{1}{n^3} \cdots = \frac{1}{n-1} $$

If it isn't then what is this called?

I haven't been able to find anything like this anywhere.

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This is a geometric series. – Akiva Weinberger Aug 31 '14 at 18:16
@columbus8myhw Thanks, but can you be a little more specific? – jonathan harding Aug 31 '14 at 18:22
up vote 2 down vote accepted

$$ if\\n\neq 1 \\ s=\frac{1}{n}+\frac{1}{n^2}+\frac{1}{n^3}+\frac{1}{n^4}+...\\multiply\\ s \\by \\n \\\frac{1}{n}s=\frac{1}{n^2}+\frac{1}{n^3}+\frac{1}{n^4}+\frac{1}{n^5}...\\now\\s-\frac{1}{n}s =\frac{1}{n}+(\frac{1}{n^2}-\frac{1}{n^2})+(\frac{1}{n^3}-\frac{1}{n^3})+(\frac{1}{n^4}-\frac{1}{n^4})+(\frac{1}{n^5}-\frac{1}{n^5})+...\\so \\s-\frac{1}{n}s =\frac{1}{n} $$ $$ s(\frac{n-1}{n})=\frac{1}{n} $$ $$ s=\frac{1}{n-1} $$

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It is a geometric series with $a=r=1/n$.

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$$\frac{1}{n}+\frac{1}{n^2}+\frac{1}{n^3}+\dots=\sum_{i=1}^{\infty} \frac{1}{n^i}=\sum_{i=1}^{\infty} \left (\frac{1}{n} \right )^i=\sum_{i=0}^{\infty} \left (\frac{1}{n} \right )^i-1=\frac{1}{1-\frac{1}{n}}-1=\frac{n}{n-1}-1=\frac{n-n+1}{n-1}=\frac{1}{n-1}$$

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Fix $r>0$ and consider $$ 1+r+r^2+r^3+\dots+r^k $$ Multiplying by $(1-r)$ you obtain $$ (1+r+r^2+r^3+\dots+r^k)(1-r)=\\ (1-r)+(r-r^2)+(r^2-r^3)+\dots+(r^{k-1}-r^k)+(r^k-r^{k+1})=\\ 1-r^{k+1}\;\;. $$ Hence you have the following identity (valid, as you can see, for $r\neq1$): $$ 1+r+r^2+r^3+\dots+r^k=\frac{1-r^{k+1}}{1-r}\;\;. $$ Then you can pass to the limit $k\to+\infty$, obtaining a finite quantity iff $|r|<1$.

This quantity clearly is $$ \frac1{1-r} $$ i.e. $$ 1+r+r^2+r^3+\dots+r^k+\dots=\frac1{1-r}\;\;. $$

Now, in order to obtain your result just pose $r=1/n$, for a fixed $n$; I guess it's a natural number, hence if $n\ge2$, clerarly $r=1/n<1$ and the above formulas are valid and easily yeld to your result.

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