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The sigma notation for the power series is equivalent to the following function: $$\sum_{n=0}^\infty x^n = \frac{1}{1-x}$$

However, given the sigma notation above, how would one arrive at it's equivalent function?

user1505087
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    For all $N\in\mathbb{N}$, we have $\sum_{n=0}^N x^n=\frac{1-x^{N+1}}{1-x}$, if $x\in]-1,1[$, then $\lim\limits_{N\rightarrow +\infty}x^N=0$ so that $\sum_{n=0}^{+\infty}x^n=\frac{1}{1-x}$. – Tuvasbien Feb 03 '20 at 21:08

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Assume that $|x|<1$, else the series does not converge. Let $S_n = \sum_{k=0}^n x^k$. Then $$ (1-x)S_n = \sum_{k=0}^n x^k - \sum_{k=0}^n x^{k+1} = 1 -x^{n+1}. $$ Dividing by $1-x$ yields $$ S_n = \frac{1 -x^{n+1}}{1-x}\stackrel{n\to\infty}\longrightarrow \frac1{1-x}. $$

Math1000
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