Prove that $1+x+x^2+...x^n=\frac{1-x^{n+1}}{1-x} $
and how to prove this implies that for $|x|<1$
$$1+x+x^2+........=\frac{1}{1-x}$$
for the first part we can prove by induction but to get second one by using first one
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Take limit $n \to \infty$ in your first part to get the answer to your second part. – User Sep 11 '17 at 14:41
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For the first part, you can use induction. You should include your attempt at the proof. – N. F. Taussig Sep 11 '17 at 14:41
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You can have a look at Prove that $1+a+a^2+\cdots+a^n=(1-a^{n+1})/(1-a)$. or Proof of the formula $1+x+x^2+x^3+ \cdots +x^n =\frac{x^{n+1}-1}{x-1}$ and other posts linked there. – Martin Sleziak Sep 11 '17 at 21:53
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Hint:
(Both for the induction part and for the limit).
Rewrite the r.h.s. as $\;\dfrac1{1-x}-\dfrac{x^{n+1}}{1-x}$.
Bernard
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