Posted by **Lisa** on Monday, July 29, 2013 at 10:34am.

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.

- Calculus-Series -
**MathMate**, Monday, July 29, 2013 at 11:08am
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_{n+1}/a_{n} < 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.

- Calculus-Series -
**Count Iblis**, Monday, July 29, 2013 at 11:39am
Integral from 1 to infinity of

(x^3)* (e^(-x^4)) dx =

1/4 exp(-1)

## Answer this Question

## Related Questions

- Calculus - For what values of p>0 does the series Riemann Sum [n=1 to ...
- Calc - Does 1/ln(x+1) converge or diverge? I've tried the nth term test, limit ...
- calculus - Does the series from 0 to infinity of [1/square root of (n+4)] x cos(...
- Calculus - Does the series (1+sin(n))/(10^n) from summation 0 to positive ...
- Calculus 2 - n=1 series to infinity (-5^n)/n^3 does it absolutely converge, ...
- Calculus - determine whether the series converges or diverges. I am stuck on ...
- calculus please help - the actual problem is integral(2 to +infinity) (1/x^2) dx...
- MATH - does the series 1/n from n=0 to infinity converge or diverge? what ...
- Calculus - does the series 1/(n^2+n) converge or diverge? n=2 to n=infinity
- calculus - is this correct? use the integral test to determine if this series is...

More Related Questions