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
On
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 $x\to \infty.$
$\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}$.