How to determine if the series is convergent or divergent.

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Puchinita5
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Homework Statement




Determine if the series is convergent or divergent.
[tex]\sum x^2e^{-x^2}[/tex]

Homework Equations





The Attempt at a Solution


[tex] x^2e^{-x^2}=\frac{x^2}{e^{x^2}}[/tex]

[tex]\lim_{x\to\infty } \frac{(x+1)^2}{e^{(x+1)^2}}\frac{e^{x^2}}{x^2}[/tex]

and since [tex](x+1)^2=x^2+2n+1[/tex]

and [tex](x^2)-(x^2+2x+1)=-(2x+1)[/tex]

I get [tex]\lim_{x\to\infty }e^{2x+1}*{(\frac{x+1}{x})}^2=\infty*1=\infty[/tex] which is [tex]> 1[/tex]


so by the root test, it is divergent.
Except I got this wrong on my exam. I was told it should be convergent. Why is this wrong?

 
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[tex] (x^2)-(x^2+2x+1)=-(2x+1)[/tex]

This is the exponent of [tex]e[/tex] on the top, so [tex]e^{2x+1}[/tex] should have been on the bottom.
 


OMG! i looked at this SO MANY TIMES and didn't see that! thank you! Oh how I love this website!
So it goes to zero, which is less than 1 and so convergent! Glad to know i was doing this right I thought i might have been WAY off!
!