Evaluate the sum of the convergent series [(-1)^n]/[(n+1)2^(2n+5)] as n goes from 0 to infinity.

Hint: Consider the function f(x) = (x^3)log(1+x^2)

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- Dec 9th 2010, 01:55 AMTaurus3sum of convergent series
Evaluate the sum of the convergent series [(-1)^n]/[(n+1)2^(2n+5)] as n goes from 0 to infinity.

Hint: Consider the function f(x) = (x^3)log(1+x^2) - Dec 9th 2010, 02:06 AMHallsofIvy
Yes, consider it! What is the MacLaurin series of that function? (You don't have to take derivatives, etc. to find that. The MacLaurin series for log(1+ x) is $\displaystyle x- x^2/2+ x^3/3+ \cdot\cdot\cdot= \sum_{n=1}^\infty (-1)^nx^n/n$. Replace that "x" with "$\displaystyle x^2$" and multiply each term by $\displaystyle x^3$.) What should you set x equal to to get the series you want?

- Dec 9th 2010, 02:12 AMTaurus3
ooh....Why thank-you for clarifying that.

- Dec 9th 2010, 02:33 AMTaurus3
hold on. Then it becomes the sum of (-1)^(n-1) [(x^5)^n]/n as n goes from 1 to infinity. In this case, the series matches the entry for ln(1+x) with x = x^5. So doesn't the sum = ln(1+x^5)? But that's weird. What do you mean by setting x equal to some value?