I have a specific problem to solve using strong law of large numbers.
Let $X_k$ be independent uniform random variables on interval $(0,k)$. Let $Y_n ={1 \over n^2}\sum\limits_{k=1}^n {X_k^3 \over k^2} $. The problem is decide if sequence $\{Y_n\}$ is absolutely convergent and if yes, find it's limit.
Theorem in the book right above the exercise is in this form : Let $\{b_n\}$ be sequence of numbers such that $0<b_1<b_2<\cdots$ and $b_n \to \infty$, let $X_k$ be indepent real random variables with finite variance and $\sum\limits_{k=1}^n \operatorname{Var}{X_k \over b_n} < \infty$ then ${1 \over b_n}\sum\limits_{k=1}^n (X_k - EX_k ) \to 0$ almost surely.
I have tried a lot of approaches but unfortunately because the mean of $X_k$ is not zero I have failed to apply the theorem. Nor I do not know, wheter the sequence really is convergent. Thanks in advance for any help/hints.
Hint: $$Y_n =m_n+{1 \over n^2}\sum\limits_{k=1}^n\xi_k,\qquad \xi_k=\frac{X_k^3-E(X_k^3)}{k^2},\qquad m_n=\frac1{n^2}\sum_{k=1}^n {E(X_k^3) \over k^2}. $$ Now, correct your condition on the variances (there is a serious typo), check that it holds for the sequence $(\xi_k)$ and $b_n=n^2$, and compute the limit of the deterministic sequence $(m_n)$.