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

In summary, the conversation is about testing the convergence of the series \sum^{∞}_{k = 1}ke^{-2k^2}. The person attempted to evaluate it in two different ways, but got different answers. They eventually discovered that the series does converge, but they are still unsure about what went wrong with the first method. After discussing their work, they realized that the u limits should have been from -2 to -∞ instead of from 1 to ∞. They corrected their mistake and were able to show that the series does converge.
  • #1
Hertz
180
8

Homework Statement



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

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?

[itex]\sum^{∞}_{k = 1}ke^{-2k^2}[/itex]

[itex]\int{^{∞}_{1}xe^{-2x^2} dx}[/itex]

Let [itex]u = -2x^2[/itex]
[itex]du = -4x dx[/itex]

[itex]\frac{-1}{4}\int{^{∞}_{1}-4xe^{-2x^2} dx}[/itex]

[itex]\frac{-1}{4}\int{^{∞}_{-2}e^{u} du}[/itex]

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

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

[itex]\frac{-1}{4}[∞ - \frac{1}{e^{2}}][/itex]

[itex]= -∞[/itex]

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

Homework Statement


Homework Equations


The Attempt at a Solution


Homework Statement


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The Attempt at a Solution


Homework Statement


Homework Equations


The Attempt at a Solution


Homework Statement


Homework Equations


The Attempt at a Solution


Homework Statement


Homework Equations


The Attempt at a Solution


Homework Statement


Homework Equations


The Attempt at a Solution


Homework Statement


Homework Equations


The Attempt at a Solution


Homework Statement


Homework Equations


The Attempt at a Solution


Homework Statement


Homework Equations


The Attempt at a Solution


Homework Statement


Homework Equations


The Attempt at a Solution

 
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  • #2
Your u limits should be -2 to MINUS infinity. Right?
 
  • #3
Dick said:
Your u limits should be -2 to MINUS infinity. Right?

They sure should. Thanks :)
 
  • #4
[itex]\int{^{∞}_{1}xe^{-2x^2} dx}[/itex]
= [itex]\int{^{∞}_{1}x/e^{2x^2} dx}[/itex]

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

=1/4[itex]\int{^{∞}_{1}1/e^{u} du}[/itex]
=1/4[itex]\int{^{∞}_{1}e^{-u} du}[/itex]

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

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

What is an improper integral?

An improper integral is an integral where one or both of the limits of integration are infinite or the integrand function has a singularity within the interval of integration.

What is the method for solving improper integrals?

The method for solving improper integrals involves taking the limit of the integral as one or both of the limits of integration approaches infinity or the singularity, and then evaluating the resulting limit.

How do you solve the improper integral for the given series, \sum^{∞}_{k = 1}ke^{-2k^2}?

The improper integral for this series can be solved using the convergence test for series with non-negative terms, which states that if the series \sum^{∞}_{k = 1}a_k converges, then the corresponding improper integral \int^{∞}_{1}a_kdx also converges.

What is the convergence test for series with non-negative terms?

The convergence test for series with non-negative terms states that if the series \sum^{∞}_{k = 1}a_k converges, then the corresponding improper integral \int^{∞}_{1}a_kdx also converges.

Is the given series, \sum^{∞}_{k = 1}ke^{-2k^2}, convergent or divergent?

The given series is convergent because it satisfies the convergence test for series with non-negative terms. Therefore, the corresponding improper integral \int^{∞}_{1}ke^{-2k^2}dx also converges.

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