Can someone please explain why ∑n=1 to ∞ ne^(-nx) converges by the integral test for x>0? Thanks!?
回答 (2)
2016-04-05 9:08 pm
✔ 最佳答案
Computing the corresponding integral:∫(t = 1 to ∞) te^(-xt) dt
= ∫(w = x to ∞) e^(-w) * dw/x, by letting w = xt, dw = x dt
= (-1/x) e^(-w) {for w = x to ∞}
= 0 - (-1/x) e^(-x)
= 1/(xe^x).
Since the integral converges, so does the corresponding series in question by the Integral Test.
I hope this helps!
2016-04-05 6:35 pm
sum ∑n=1 to ∞ ne^(-nx)
= ∑n=1 to ∞ nu^n ,,,,,,,,,,,, where u = e^-x
= 1/(1-e^-x)^2 see below
since the sum of [1 to oo ] 1 + 2x + 3x^2 + 4x^3 + ................ = 1/(1-x)^2
see https://www.algebra.com/algebra/homework/Sequences-and-series/Sequences-and-series.faq.question.117354.html
= ∑n=1 to ∞ nu^n ,,,,,,,,,,,, where u = e^-x
= 1/(1-e^-x)^2 see below
since the sum of [1 to oo ] 1 + 2x + 3x^2 + 4x^3 + ................ = 1/(1-x)^2
see https://www.algebra.com/algebra/homework/Sequences-and-series/Sequences-and-series.faq.question.117354.html
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