Answer
\[\text{Converges}\]
Work Step by Step
\[\begin{align}
& \sum\limits_{k=2}^{\infty }{\frac{1}{{{e}^{k}}}} \\
& \text{ }{{a}_{k}}=\frac{1}{{{e}^{k}}}={{e}^{-k}},\text{ then let }f\left( x \right)={{e}^{-x}} \\
& \text{We know that }f\left( x \right)={{e}^{-x}}\text{ is continuous and positive for }x\ge 1, \\
& \text{now we will determine if }f\left( x \right)\text{ is decreasing}\text{, then } \\
& f\left( x \right)={{e}^{-x}} \\
& f'\left( x \right)=-{{e}^{-x}} \\
& \text{For all }x\ge 1\text{ }f'\left( x \right)<0,\text{ then the function is decreasing and} \\
& \text{satisfies the conditions for the integral test}\text{.} \\
& \int_{2}^{\infty }{{{e}^{-x}}}dx \\
& \text{Solving improper integral} \\
& \int_{2}^{\infty }{{{e}^{-x}}}dx=\underset{b\to \infty }{\mathop{\lim }}\,\int_{2}^{b}{{{e}^{-x}}}dx \\
& \text{ }=\underset{b\to \infty }{\mathop{\lim }}\,\left[ -{{e}^{-x}} \right]_{2}^{b} \\
& \text{ }=\underset{b\to \infty }{\mathop{\lim }}\,\left[ -{{e}^{-b}}+{{e}^{-2}} \right] \\
& \text{ }=-\underset{b\to \infty }{\mathop{\lim }}\,\left[ {{e}^{-b}} \right]+\underset{b\to \infty }{\mathop{\lim }}\,\left[ {{e}^{-2}} \right] \\
& \text{ }=0+{{e}^{-2}} \\
& \text{ }=\frac{1}{{{e}^{2}}} \\
& \text{The integral converges}\text{, so by the Integral test }\left( \text{Theorem 8}\text{.10} \right) \\
& \text{the series converges} \\
\end{align}\]