1. ## Integral Test

Summation, n=1 to infinity (ne^(-n/2))

2. Hello, saiyanmx89!

Integral test: .$\displaystyle \sum^{\infty}_{n=1} ne^{\text{-}\frac{n}{2}}$

The corresponding integral is: .$\displaystyle \int^{\infty}_1 xe^{\text{-}\frac{1}{2}x}dx$

By parts: . $\displaystyle \begin{array}{ccccccc}u &=& x & & dv &=& e^{\text{-}\frac{1}{2}x}dx \\ du &=& dx && v &=& -2e^{\text{-}\frac{1}{2}x} \end{array}$

We have: .$\displaystyle -2xe^{\text{-}\frac{1}{2}x} + 2\int e^{\text{-}\frac{1}{2}x}dx \;=\;-2xe^{\text{-}\frac{1}{2}x} - 4e^{\text{-}\frac{1}{2}x} \;=\;\frac{-2(x+2)}{e^{\frac{x}{2}}}\,\bigg]^{\infty}_1$

Then: .$\displaystyle \lim_{b\to\infty}\frac{-2(x+2)}{e^{\frac{x}{2}}}\,\bigg]^b_1 \;=\;\lim_{b\to\infty}\bigg[\frac{-2(b+2)}{e^{\frac{b}{2}}} + \frac{2(2)}{e^{\frac{1}{2}}} \bigg] \;\;=\;\;0 + \frac{4}{e^{\frac{1}{2}}} \;\;=\;\;\frac{4}{\sqrt{e}}$

The integral converges . . . Therefore, the series will converge.

3. #1 - Integrate the function.

You should arrive at -2x*e^(-x/2) -4e^(-x/2).

Now evaluate the integral for the bounds lim N->infinity (1, N).