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For $n\geq 1$ and $x \gt 0$, define

$$ R_n(x)=\frac{1}{n}\sum_{k=1}^{n}\frac{1}{(x+\frac{2k-1}{2n})^2} $$

then $R_n(x)$ is a Riemann sum, which converges to the integral

$$ R=\int_{0}^1 \frac{dt}{(x+t)^2}=\frac{1}{x}-\frac{1}{x+1} $$

We have

$$ R-R_1(x)=\frac{1}{x(x+1)(2x+1)^2} $$

$$ R-R_2(x)=\frac{16x^2+16x+9}{x(x+1)(4x+1)^2(4x+3)^2} $$

$$ R-R_3(x)=\frac{3888x^4 + 7776x^3 + 6984x^2 + 3096x + 675}{x(x+1)(6x+1)^2(6x+3)^2(6x+5)^2} $$

Thus for $n\leq 3$, the numerator of $R-R_n(x)$ has positive coefficients. Can anyone show that $R>R_n(x)$ for any $n$ and $x$, by showing those coefficients are always positive, or by any other method?

By the way, this might look like homework, but it isn't.

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Maybe investigate this: When a function is convex, is there an inequality between the integral and the midpoint Riemann sum for the integral? – GEdgar May 5 '12 at 15:49
As mentioned by GEdgar, the geometry takes care of things. – André Nicolas May 5 '12 at 16:07
GEdgar, André : thanks for answering my stupid question. – Ewan Delanoy May 5 '12 at 17:33
up vote 1 down vote accepted

The comments on the OP say it all. Just to make things a little more explicit : if $f$ is a convex, integrable function $[a,b] \to \mathbb R$, then we have $f\left(\frac{a+b}{2}\right) \leq \frac{f(t)+f(a+b-t)}{2}$ for any $t \in [a,b]$. Integrating,we deduce $\int_a^b f(t)dt \geq (b-a)f\left(\frac{a+b}{2}\right)$. Thus, in the mid-point Riemann sum

$$ R_n(f,x)=\frac{1}{n}\sum_{k=1}^{n}f\left(x+\frac{2k-1}{2n}\right) $$

each individual term $\frac{1}{n}\cdot f\left(x+\frac{2k-1}{2n}\right)$ is lower than $\int_{\frac{2k-2}{2n}}^{\frac{2k}{2n}} f(t)dt$, and summing we obtain

$$ R_n(f,x) \leq R=\int_{0}^{1} f(t)dt $$

The original question is about $f(x)=\frac{1}{x^2}$.

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