Answer
Function satisfies conditions for integral test
Series converges
Work Step by Step
In order integral test can be applied $\sum ^{\infty }_{n=1}a_{n}$
$f(x) $ must be positive ,continuous and decreasing for $x \geq 1$
$\sum ^{\infty }_{n=1}ne^{-n/2}=e^{-\dfrac {1}{2}}+\sum ^{\infty }_{n=2}ne^{-\dfrac {n}{2}}$
We see that $f\left( x\right) =xe^{-\frac {x}{2}}$ is positive and continuous for $x \geq2$ and in order to find if the function is decreasing we need to find derivative of function
$f'\left( x\right) =\left( 1-\dfrac {x}{2}\right) e^{-\dfrac {x}{2}}$
For $x>2$
the function is decreasing
So it satisfies all three conditions for integral test
To find whether series is converges or diverges, we see whether the integral converges or diverges.
$$ \int ^{\infty }_{1}xe^{-\frac {x}{2}}=\lim _{b\rightarrow \infty }2e^{-\frac {x}{2}}\left( x+2\right) \int ^{b}_{1}=\lim _{b\rightarrow \infty }2\left( e^{-\frac {b}{2}}\left( b+2\right) -3xe^{-\frac
{1}{2}}\right) $$
$\lim _{b\rightarrow \infty }\dfrac {\left( b+2\right) }{e^{-\dfrac {b}{2}}}=\dfrac {\dfrac {d}{db}\left( b+2\right) }{\dfrac {d}{db}\left( e^{-\dfrac {b}{2}}\right) }=\dfrac {1}{-\dfrac {e^{-\dfrac {b}{2}}}{2}}=0\Rightarrow \int ^{\infty }_{1}xe^{-\dfrac {x}{2}}=6e^{-\dfrac {1}{2}}$
The integral converges, so the series converges