Question
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
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.
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)
Related Questions
Does the series (1+sin(n))/(10^n) from summation 0 to positive infinity converge or diverge?
I le...
n=1 series to infinity (-5^n)/n^3 does it absolutely converge, diverge or conditionally converge. Wo...
Consider the infinite geometric series below.
a. Write the first 4 terms of the series
b. Does the...
Consider the infinite geometric series infinity sigma n=1 -4(1/3)^n-1. In this, the lower limit of t...