Asked by Lisa
Sigma with n=1 to n= positive infinity
(x^3)* (e^(-x^4))
Does it converge or diverge and use the integral test to show how. I got that it diverges, but this isn't correct. Can someone please explain? Thanks.
(x^3)* (e^(-x^4))
Does it converge or diverge and use the integral test to show how. I got that it diverges, but this isn't correct. Can someone please explain? Thanks.
Answers
Answered by
MathMate
I assume the expression is
(n^3)* (e^(-n^4)) for n=1 to +&inf;
You can use the ratio test to show that it converges, namely, show that
a<sub>n+1</sub>/a<sub>n</sub> < 1 ∀ k≥1.
The integral test will result in a definite integral from 1 to +&inf; which is not an elementary function, but evaluates numerically to about 0.09. However, this still shows that the series is convergent.
(n^3)* (e^(-n^4)) for n=1 to +&inf;
You can use the ratio test to show that it converges, namely, show that
a<sub>n+1</sub>/a<sub>n</sub> < 1 ∀ k≥1.
The integral test will result in a definite integral from 1 to +&inf; which is not an elementary function, but evaluates numerically to about 0.09. However, this still shows that the series is convergent.
Answered by
Count Iblis
Integral from 1 to infinity of
(x^3)* (e^(-x^4)) dx =
1/4 exp(-1)
(x^3)* (e^(-x^4)) dx =
1/4 exp(-1)
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