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I was playing around with fractions in my head when I realized the infinite summation of $\frac{1}{b^n}=\frac{1}{(b-1)}$ for when $b>1$.

I tried writing a proof for this and researching the function, but found nothing, so does anyone have any insights to how I can prove this? Thanks in advance.

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I assume you sum up from $n=1$ to $\infty$. This is a very well-known result. For each $q\in\mathbb C$ with $|q|<1$ holds $$ \sum_{n=0}^\infty q^n=\frac1{1-q}. $$ The series on the LHS is called geometric series. If you google proof geometric series you find some.

Now if you choose $b>1$ and $q=\frac1b$ you get $|q|=\frac1b<1$ and $$ \sum_{n=1}^\infty \frac1{b^n}=\sum_{n=1}^\infty q^n = \frac1{1-q}-1=\frac{b}{b-1}-1=\frac1{b-1} $$

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