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In my lecture notes about proof of strong law of large number. The last step to show $\sum_n^\infty \frac{var(Y_n)}{n^2}$ converges is $$E(X^2\sum_{n>max\{|X|,1\}}^\infty\frac{2}{n(n+1)} )\leq 2E(|X|+1)$$ I referred other books, one of them only shows it is less than or equal $2E(|X|)$. I do not know how to get it. Can anyone show me? Thanks.

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Observe first that \begin{equation*}\sum\limits_{n=m}^\infty\frac{1}{n(n+1)}=\lim\limits_{N\to\infty}\sum\limits_{n=m}^N \frac{1}{n(n+1)}=\lim\limits_{N\to\infty}\sum\limits_{n=m}^N \left(\frac{1}{n}-\frac{1}{n+1}\right) =\lim\limits_{N\to\infty}\left(\frac{1}{m}-\frac{1}{N+1}\right)=\frac{1}{m}.\end{equation*} Thus \begin{equation*} \sum\limits_{n>\max\{|X|,1\}}\frac{1}{n(n+1)}\leqslant\frac{1}{\lfloor|X|\rfloor+1}\mathbf{1}_{[1,\infty)}(|X|)+\tfrac{1}{2}\mathbf{1}_{[0,1)}(|X|).\end{equation*} Hence \begin{align*} X^2\sum\limits_{n>\max\{|X|,1\}}\frac{2}{n(n+1)}&\leqslant \frac{2X^2}{\lfloor |X|\rfloor+1}\mathbf{1}_{[1,\infty)}(|X|)+X^2\mathbf{1}_{[0,1]}(|X|)\\ &\leqslant \frac{2X^2}{|X|}\mathbf{1}_{[1,\infty)}(|X|)+X^2\mathbf{1}_{[0,1]}(|X|)\\ &=2|X|\mathbf{1}_{[1,\infty)}(|X|)+X^2\mathbf{1}_{[0,1]}(|X|),\end{align*} and the expectation of the last sum is no more than $2E(|X|+1)$.

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