# Limit of $\sum\limits_{i=1}^\infty \frac{1}{a^i}$ as x -> infinity

I observed something while working a bit about series.

I found out that the limit of :

$$\sum\limits_{i=1}^x \frac{1}{a^i}$$

as x approches infinity seems to be equal to $\frac{1}{a-1}$.

If this formula is true, we can prove that :

$$\lim\limits_{x \to \infty} \sum\limits_{i=1}^x \frac{1}{e^i} = \frac{1}{e-1} \approx 0.58198$$

How can we prove this formula ?

Edit : fixed no x

• en.wikipedia.org/wiki/Geometric_series – Crostul Jul 2 '16 at 13:04
• You have no $x$ in your formula. – lisyarus Jul 2 '16 at 13:07
• Perhaps you mean $\sum_{i=1}^\infty 1/e^i\equiv\lim_{x\rightarrow\infty}\sum_{i=1}^x 1/e^i$ – parsiad Jul 2 '16 at 13:07
• It might help to note that $$\frac1{a-1}=\frac1a\cdot\frac1{1-\frac1a}.$$ – Did Jul 2 '16 at 13:13

$\sum_{i=1}^{n}$$\frac{1}{a^i} =\frac{1}{a}+\frac{1}{a^2}+ \frac{1}{a^2}+..... which is an infinite geometric progression valid for \frac{1}{|a|}<1 or |a|>1. \implies \sum_{i=1}^{n}$$\frac{1}{a^i}=$$\frac{1/a}{1-1/a}=\frac{1}{a-1}. For 0\ne a\ne 1 and positive integer x we have$$\sum_{i=1}^x a^{-i}=a^{-1}(1-a^{-x-1})/(1-a)\text { because }\; (1-a)\sum_{i=1}^xa^{-i}=1-a^{-x-1}.$$For |a|>1 we have \lim_{x\to \infty}a^{-x-1}=0 so for |a|>1 the summation converges to$$a^{-1}/(1-a^{-1})=1/(a-1).$$Remark: For |a|>1, let |a|=1+b with b>0. The statement S(x): |a|^x\geq 1+b is true when x=1. If S(x) is true for some x\geq 1 then$$|a|^{1+x}=|a|^x\cdot |a|\geq (1+xb)|a|=(1+xb)(1+b)=1+(1+x)b+xb^2>1+(x+1)b.$$So$S(x)\implies S(1+x)$for$x\geq 1.$By induction on$x$we have$|a|^x\geq 1+xb$and hence$0<|a|^{-x}<1/(1+xb) ,\;. $So$a^{-x}$converges to$0$as$x\to \infty.\$