What is the limit of $\sum _{k=0}^n\:\frac{\exp\left(-k\right)}{2^k}$ when $n->\infty $?

Question

I just wanna know what is the limit of this serie above. I proved that this serie is convergent using Cauchy's convergence test, and this serie is increasing, and $\forall n\in \mathbb{N},\:Un<=2$ So the limit must be 2. Am I right ?

Answer 1

That is a geometric series with ratio $\frac 1 {2e}$. As $n\to \infty$, it converges to $\frac 1{1-\frac {1}{2e}}$, or around 1.2253996735605641.

Answer 2

$$ 2^k = \mathrm{e}^{k\ln 2} $$ so you are computing the sum $$ \sum_{k=0}^n\mathrm{e}^{-k-k\ln 2}=\sum_{k=0}^n\mathrm{e}^{-(1+\ln 2)k} $$ or $$ \sum_{k=0}^n\left[\mathrm{e}^{-(1+\ln 2)}\right]^k $$ which we can state as $a^k$ thus we have $$ \sum_{k=0}^n a^k $$ this is a geometric sum so as long as $$ a = \mathrm{e}^{-(1+\ln 2)} < 1 $$ we can converge.