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)
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?
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?