Find $E[(1-X)^k]$ given $E[X^k] = \frac{1}{(k+1)^2}$
The previous part of the question involved finding $E[X^k]$ given $X=AB$, where $A$ and $B$ are uniformly distributed over $[0, 1]$. However, I don't think this helps with the matter at hand.
My intuition was to expand $E[(1-X)^k]$ as a series, which gave me a nasty sum I couldn't compress. Is there a more direct approach? Can I avoid using $A$ and $B$?
Define $$u_n:=(n+1)E(1-X)^n=(n+1)\sum_{k=0}^n\binom{n}{k}\frac{(-1)^k}{(k+1)^2}=\sum_{k=0}^n\binom{n+1}{k+1}\frac{(-1)^k}{k+1}.$$We'll prove this is a displaced harmonic sequence, viz. $u_n=H_{n+1}$. Clearly $u_0=1$ as required, so we only need verify $u_{n+1}-u_n=\frac{1}{n+2}$. Indeed, this difference is $$\frac{(-1)^{n+1}}{n+2}+\sum_{k=0}^{n}\binom{n+1}{k}\frac{(-1)^k}{k+1}=\frac{(-1)^{n+1}}{n+2}+\frac{1}{n+2}\sum_{k=0}^{n}(-1)^k\binom{n+2}{k+1}\\=\frac{(-1)^{n+1}}{n+2}+\frac{1}{n+2}\left(1+(-1)^n-\color{blue}{\sum_{j=0}^{n+2}(-1)^j\binom{n+2}{j}}\right).$$The blue term is $(1-1)^{n+2}=0$ by the binomial theorem, so we have $$u_{n+1}-u_n=\frac{1+(-1)^n+(-1)^{n+1}}{n+2}=\frac{1}{n+2}.$$