# How do I go from a summation form to integral form of an expression?

How is the following inequality valid?

$$\sum\limits_{j = 1}^{N/2} \sum\limits_{n = j}^{\infty} \frac{1}{n^2} < \sum_{j = 1}^{N/2} \int_{j - 1/2}^{\infty} \frac{dx}{x^2}$$.

I came across this inequality in one of the references I'm using, and the author states the above is true because $1/x^2$ is a strictly convex function on $(0, \infty)$.

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Note that since $\dfrac{1}{x^2}$ is a decreasing function, $\int_{j-1}^j\frac{\mathrm{d}x}{x^2}>\frac{1}{j^2}$. Therefore, $$\int_{j-1}^\infty\frac{\mathrm{d}x}{x^2}>\sum_{n=j}^\infty\frac{1}{n^2}\tag{1}$$ However to get the inequality cited in the question, we need to evaluate a bit more carefully: \begin{align} \int_{j-1/2}^{j+1/2}\frac{\mathrm{d}x}{x^2} &=\frac{1}{j-1/2}-\frac{1}{j+1/2}\\ &=\frac{1}{j^2-1/4}\\ &>\frac{1}{j^2}\tag{2} \end{align} Summing $(2)$ yields $$\int_{j-1/2}^\infty\frac{\mathrm{d}x}{x^2}>\sum_{n=j}^\infty\frac{1}{n^2}\tag{3}$$ Using the strict convexity of $\mathbf{\dfrac{1}{x^2}}$
We can prove $(3)$ using strict convexity. Since $x$ is not constant on the support of uniform measure ($\mathrm{d}x$) on $[j{-}\frac12,j{+}\frac12]$, Jensen's inequality yields $$\int_{j-1/2}^{j+1/2}\frac{\mathrm{d}x}{x^2}>\frac{1}{\left(\int_{j-1/2}^{j+1/2}x\,\mathrm{d}x\right)^2}=\frac{1}{j^2}$$