Does the series from k = 1 to infinity k^3 e^-k converge? I think it diverges, but I'm not sure why.?
回答 (1)
2015-02-02 5:48 am
✔ 最佳答案
Using the Ratio Test,r = lim(k→∞) ((k+1)^3 e^(-(k+1))) / (k^3 e^(-k))
..= lim(k→∞) e^(-1) * (k+1)^3/k^3
..= lim(k→∞) e^(-1) * ((k+1)/k)^3
..= lim(k→∞) e^(-1) * (1 + 1/k)^3
..= e^(-1) * (1 + 0)^3
..= 1/e.
Since r = 1/e < 1, the series converges.
I hope this helps!
收錄日期: 2021-04-21 01:16:31
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