Integral Test

**saiyanmx89** · March 31st 2009, 02:07 PM

Does this fail the Integral Test? I think it does.

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

**TheEmptySet** · March 31st 2009, 02:43 PM

Originally Posted by saiyanmx89

Does this fail the Integral Test? I think it does.

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

$\int_{1}^{\infty}xe^{-\frac{x}{2}}dx$

using integration by parts ( and L'hosiptals rule)you get

and please forgive my abuse of notation

$\int_{1}^{\infty}xe^{-\frac{x}{2}}dx=-\frac{x}{2}e^{-\frac{x}{2}}\bigg|_{1}^{\infty}-\int_{1}^{\infty}-\frac{1}{2}e^{-\frac{x}{2}}dx=\frac{e^{-\frac{1}{2}}}{2}+\frac{1}{2}\int_{1}^{\infty}e^{-\frac{x}{2}}dx$

Since the latter integral also coverges the series coverges by the integral test.

**Soroban** · March 31st 2009, 02:54 PM

Helo, saiyanmx89!

Does this fail the Integral Test? I think it does. . $\sum^{\infty}_{n=1} n\,e^{\text{-}\frac{n}{2}}$

I believe it passes . . .

We have: . $\int^{\infty}_1 x\,e^{\text{-}\frac{x}{2}}dx$

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

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

Then: . $\lim_{b\to\infty}\bigg[\frac{-2(x+2)}{e^{\frac{x}{2}}}\,\bigg]^b_1 \;=\;\lim_{b\to\infty}\bigg[\frac{-2(b+2)}{e^{\frac{b}{2}}} + \frac{6}{e^{\frac{1}{2}}}\bigg] \;=\;0 + \frac{6}{e^{\frac{1}{2}}}$

Therefore, the integral converges to: . $\frac{6}{\sqrt{e}}$

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