36. ∑(-e)^(-k) from k = 1 to ∞

Explanation:

Step1: Identify the geometric - series form

A geometric series has the form $\sum_{k = 1}^{\infty}ar^{k - 1}$. Rewrite the given series $\sum_{k = 1}^{\infty}(-e)^{-k}=\sum_{k = 1}^{\infty}\frac{1}{(-e)^{k}}=\sum_{k = 1}^{\infty}\frac{1}{-e}\cdot(\frac{1}{-e})^{k - 1}$, where $a=\frac{1}{-e}$ and $r=\frac{1}{-e}$.

Step2: Check the convergence condition

For a geometric series $\sum_{k = 1}^{\infty}ar^{k - 1}$ to converge, $|r|\lt1$. Here, $r =-\frac{1}{e}$, and $|r|=\left|-\frac{1}{e}\right|=\frac{1}{e}\approx0.368\lt1$.

Step3: Use the formula for the sum of an infinite geometric series

The sum of an infinite geometric series $\sum_{k = 1}^{\infty}ar^{k - 1}=\frac{a}{1 - r}$ when $|r|\lt1$. Substituting $a =-\frac{1}{e}$ and $r=-\frac{1}{e}$ into the formula, we get $S=\frac{-\frac{1}{e}}{1-(-\frac{1}{e})}$.

Step4: Simplify the expression

First, simplify the denominator: $1-(-\frac{1}{e})=1 + \frac{1}{e}=\frac{e + 1}{e}$. Then, $S=\frac{-\frac{1}{e}}{\frac{e + 1}{e}}=-\frac{1}{e + 1}$.

Answer:

$-\frac{1}{e + 1}$