Solving Improper Integral: \sum^{∞}_{k = 1}ke^{-2k^2}

[Hertz]

Homework Statement

I'm trying to test whether the sequence converges or not:
$\sum^{∞}_{k = 1}ke^{-2k^2}$

2. The attempt at a solution

I tried to evaluate this in two ways, each of which produced different answers. I was able to eventually discover that this series does converge, but I still don't see what was wrong with the first method I tried (which told me it diverged.)

Could someone please take a look at my work and tell me what I did wrong?

$\sum^{∞}_{k = 1}ke^{-2k^2}$

$\int{^{∞}_{1}xe^{-2x^2} dx}$

Let $u = -2x^2$
$du = -4x dx$

$\frac{-1}{4}\int{^{∞}_{1}-4xe^{-2x^2} dx}$

$\frac{-1}{4}\int{^{∞}_{-2}e^{u} du}$

$\frac{-1}{4}{lim}_{b → ∞}[e^u]^{b}_{-2}$

$\frac{-1}{4}[{lim}_{b → ∞}(e^b) - \frac{1}{e^{2}}]$

$\frac{-1}{4}[∞ - \frac{1}{e^{2}}]$

$= -∞$

However, if you instead let $u = 2x^{2}$ it can be shown that the series converges. (Along with the integral)

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The Attempt at a Solution

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[Dick]

Your u limits should be -2 to MINUS infinity. Right?

 

[Hertz]

Your u limits should be -2 to MINUS infinity. Right?

They sure should. Thanks :)

 

[STEMucator]

$\int{^{∞}_{1}xe^{-2x^2} dx}$
= $\int{^{∞}_{1}x/e^{2x^2} dx}$

u=2x^2
1/4du = xdx

=1/4$\int{^{∞}_{1}1/e^{u} du}$
=1/4$\int{^{∞}_{1}e^{-u} du}$

Integrate that, sub back in for u, take the limit, and you should be done.