I think that's true but I don't know how I can prove it logically.

Question

Firstly I got it in my exam so I didn't know how I can prove that this sum cannot equal 2 or more.

By: $0\lt a \lt 1$

$\sum_{k=0}^n\ (a^k) \lt 2$

Answer 1

You can't because it's not true. Take $a=\frac{2}{3}$, and $n=2$. \begin{align*} \sum_{k=0}^na_k&=\sum_{k=0}^2 \left(\frac{2}{3}\right)^k\\ &=1+\frac{2}{3}+\frac{4}{9}\\ &=\frac{19}{9}\\ &>2 \end{align*}

You tagged this with Cauchy Sequences; what you could do is notice that your sum is the $n^{th}$ partial sum for the sequence $a^k$. Since $0<a<1$, this sequence of partial sums is convergent, and therefore Cauchy. This implies the differences between terms gets small, so you could get $\left | \sum_{k=0}^na^k - \sum_{k=0}^ma^k\right|<2 $ for $n,m>N$ for sufficiently large $N$.