Use the integral test to find the sum of the series from 1 to infinity: ne^(-n^2)

Question

I found the integral of xe^-x^2 to be -1/2 (e^(-x^2)) and am having trouble evaluating the limit as t goes to infinity of -1/2 e^(-t^2) +1/(2e)

Answer 1

Hello Christina,

You can indeed check the given series for convergence using the integral test. Here, you will evaluate the improper integral ∫[1 to ∞] xe^{-x^2} dx. This integral is defined as

∫[1 to ∞] xe^{-x^2} dx = lim(t→∞) ∫[1 to t] xe^{-x^2} dx. You carried out the integration correctly and obtained

∫[1 to ∞] xe^{-x^2} dx = lim(t→∞) ∫[1 to t] xe^{-x^2} dx.

=lim(t→∞) {[(-1/2)e^{-x^2}](evaluate from 1 to t]}

=lim(t→∞) {(-1/2)e^{-t^2} - (-1/2)e^^{(-1)^2}]} = lim(t→∞) {(-1/2)e^{-t^2} +1/(2e)}

Now for the first term in the above limit, note that (-1/2)e^{-t^2} = -1/(2e^{t^2}), and since e^{t^2} → ∞ as t→∞, we have

lim(t→∞) (-1/2)e^{-t^2} = 0.

Therefore,

∫[1 to ∞] xe^{-x^2} dx = lim(t→∞) {(-1/2)e^{-t^2} +1/(2e)} = 1/(2e).

Your result for the integral is correct! Thus, by the integral test, we conclude that the given infinite series converges, since the associated improper integral converges (equals a real number).

By the way, from the statement of your question I'm not sure if you were supposed to attempt to find the sum of the series. The Integral Test allows us to determine whether the infinite series converges or diverges, but does not tell us the sum of the series.

Hope this helps. Let me know if you have any more questions.

William