Answer
\[\text{The series converges}\]
Work Step by Step
\[\begin{align}
& \sum\limits_{k=1}^{\infty }{k{{e}^{-2{{k}^{2}}}}} \\
& \sum\limits_{k=1}^{\infty }{{{a}_{k}}\Rightarrow }\text{ }{{a}_{k}}=k{{e}^{-2{{k}^{2}}}},\text{ then let }f\left( x \right)=x{{e}^{-2{{x}^{2}}}} \\
& \text{We know that }f\left( x \right)=x{{e}^{-2{{x}^{2}}}}\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)=x{{e}^{-2{{x}^{2}}}} \\
& \text{Differentiating by the product rule} \\
& f'\left( x \right)=x\left( -4x{{e}^{-2{{x}^{2}}}} \right)+{{e}^{-2{{x}^{2}}}} \\
& f'\left( x \right)={{e}^{-2{{x}^{2}}}}\left( 1-4{{x}^{2}} \right) \\
& f'\left( x \right)=0 \\
& 1-4{{x}^{2}}=0\to {{x}_{1}}=-\frac{1}{2},\text{ }x=\frac{1}{2} \\
& \text{Test }f'\left( x \right)\text{ for the interval }\left( 1,\infty \right): \\
& f'\left( 2 \right)={{e}^{-2{{\left( 2 \right)}^{2}}}}\left( 1-4{{\left( 2 \right)}^{2}} \right)<0,\text{ decreasing}\text{.} \\
& \text{The function satisfies the conditions for the integral test}\text{.} \\
& \\
& \int_{1}^{\infty }{x{{e}^{-2{{x}^{2}}}}}dx \\
& \text{Solving improper integral} \\
& \int_{1}^{\infty }{x{{e}^{-2{{x}^{2}}}}}dx=\underset{b\to \infty }{\mathop{\lim }}\,\int_{1}^{b}{x{{e}^{-2{{x}^{2}}}}}dx \\
& \text{ }=-\frac{1}{4}\underset{b\to \infty }{\mathop{\lim }}\,\int_{1}^{b}{{{e}^{-2{{x}^{2}}}}\left( -4x \right)}dx \\
& \text{ }=-\frac{1}{4}\underset{b\to \infty }{\mathop{\lim }}\,\left[ {{e}^{-2{{x}^{2}}}} \right]_{1}^{b} \\
& \text{ }=-\frac{1}{4}\underset{b\to \infty }{\mathop{\lim }}\,\left[ {{e}^{-2{{\left( b \right)}^{2}}}}-{{e}^{-2}} \right] \\
& \text{Evaluate the limit when }b\to \infty \\
& \text{ }=-\frac{1}{4}\underset{b\to \infty }{\mathop{\lim }}\,\left[ {{e}^{-2{{\left( b \right)}^{2}}}} \right]+\frac{1}{4}\underset{b\to \infty }{\mathop{\lim }}\,\left[ {{e}^{-2}} \right] \\
& \text{ }=0+\frac{1}{4{{e}^{2}}} \\
& \text{ }=\frac{1}{4{{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}\]