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I am working though YS Chow's Probability Theory and have found an equality that I cannot justify. In theorem 3 on page 118 the following inequality is used:

$\sum_{n=j}^\infty \frac{1}{n^2} \le \frac{1}{j^2} + \frac{1}{j}$

I feel like this should very elementary to show but I can't seem to work it out or find any references.

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Use the fact that $k^2 > k(k-1)$. Hence, we have $$\dfrac1{k^2} < \dfrac1{k(k-1)}$$ This gives us \begin{align} \sum_{n=j}^{\infty} \dfrac1{n^2} & = \dfrac1{j^2} + \sum_{n=j+1}^{\infty} \dfrac1{n^2} < \dfrac1{j^2} + \sum_{n=j+1}^{\infty} \dfrac1{n(n-1)}\\ & = \dfrac1{j^2} + \sum_{n=j+1}^{\infty} \left(\dfrac1{n-1} - \dfrac1n \right)\\ & = \dfrac1{j^2} + \dfrac1{j} \end{align}

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