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The inequality itself is listed under summation approximations via integrals. It is being applied to a monotonically increasing function $f:\mathbb{R} \rightarrow \mathbb{R}$:

$$\int_{a-1}^{b} f(s)\ ds \le \sum_{i=a}^{b} f(i) \le \int_{a}^{b+1} f(s)\ ds$$

After looking at the Euler–Maclaurin formula article, I believe that the inequality is some result of the approximation of the integral $$\int_{a}^{b} f(s)\ ds$$ by $f(a) + f(a+1) + \ldots + f(b)$. Therefore, it would make sense that for a monotonically increasing function $f$,

$$ \int_{a-1}^{b} f(s)\ ds \le \int_{a}^{b} f(s)\ ds \le \int_{a}^{b+1} f(s)\ ds \\ \Rightarrow \quad \int_{a-1}^{b} f(s)\ ds \le \sum_{i=a}^{b} f(i) \le \int_{a}^{b+1} f(s)\ ds. $$

This, however, is a very sloppy justification. Any direction towards a more formal approach would be greatly appreciated!

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You have $$\int_{a-1}^b f(s) ds = \sum_{i=a}^{b} \int_{i-1}^{i} f(s) ds$$

But, as $f$ is increasing,

$$\forall i, \ \int_{i-1}^{i} f(s) ds \leq \int_{i-1}^{i} f(i) ds = f(i)$$

So,

$$\int_{a-1}^b f(s) ds \leq \sum_{i=a}^{b} f(i)$$

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This diagram illustrates that $$ \int_0^9f(x)\,\mathrm{d}x\le\sum_{n=1}^9f(n)\tag1 $$ which follows from summing $$ \int_{n-1}^nf(x)\,\mathrm{d}x\le f(n)\tag2 $$ which is true because $f(x)\le f(n)$ for $x\in[n-1,n]$.


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This diagram illustrates that $$ \sum_{n=1}^9f(n)\le\int_1^{10}f(x)\,\mathrm{d}x\tag3 $$ which follows from summing $$ f(n)\le\int_n^{n+1}f(x)\,\mathrm{d}x\tag4 $$ which is true because $f(n)\le f(x)$ for $x\in[n,n+1]$.

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  • $\begingroup$ Nice drawings +1. $\endgroup$ – Paramanand Singh Jun 15 '18 at 14:25
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For example $$ \int_a^{b+1} f(x)\, dx=\sum_{i=a}^b\int_{i}^{i+1}f(x)\, dx\ge \sum_{i=a}^b f(i) $$ as $$ \int_{i}^{i+1}f(x)\ge \int_{i}^{i+1}f(i)=f(i) $$ since $f$ is increasing.

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Note that $$f(i)=\int_{i}^{i+1} f(i)\ ds\leq \int_{i}^{i+1} f(s)\ ds$$ and that $$f(i)=\int_{i-1}^{i} f(i)\ ds\geq \int_{i-1}^{i} f(s)\ ds$$

Using these inequalities in the sum, you will find the desired result.

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$\sum _{i=a} ^bf(i)$ is an inferior sum of the integral $\int _a ^{b+1}f(s)ds$ and a superior sum of $\int _{a-1} ^b f(s)ds$.

This is a consecuence of the monotony of $f$. If you partition the interval $[a-1,b]$ taking $t_i = (a-1)+1$ we obtain

$$\qquad t_{i+1} - t_i = 1 \qquad \text{ and } \qquad M_i = \sup \{f(x) : x\in [t_i,t_{i+1}] \} = f(t_i)$$

Then, taking $n=b-(a-1) = b-a +1$ $$\int _ {a-1} ^b f(s)ds \leq \sum _{i=1} ^n M_i(t_{i+1} - t_{i}) = \sum _{i=1} ^n f(t_{i+1}) = \sum _{i=1} ^nf(a+i) = \sum _{j=a} ^b f(j)$$ For the last equality you make the change $j = (a-1) + i$ so for $i=1,\; j=a$ and for $i=b-a+1$ you get $j=b$

The other inequality is analogus considering that $f(t_{i-1})$ is the infimum of the $f(x)$ for $x\in [t_i, t_{i+1}]$ and taking the interval $[a,b+1]$ in consideration.

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Note that $f(i)-\int_i^{i+1} f(x) dx =\int_i^{i+1} (f(i)-f(x)) dx $ and $f(i+1)-\int_i^{i+1} f(x) dx =\int_i^{i+1} (f(i+1)-f(x)) dx $.

If $f$ is increasing then $f(i)-f(x) \le 0$ and $f(i+1)-f(x) \ge 0$ so $f(i)-\int_i^{i+1} f(x) dx \le 0$ and $f(i+1)-\int_i^{i+1} f(x) dx \ge 0$.

Similarly, if $f$ is decreasing then $f(i)-f(x) \ge 0$ and $f(i+1)-f(x) \le 0$ so $f(i)-\int_i^{i+1} f(x) dx \ge 0$ and $f(i+1)-\int_i^{i+1} f(x) dx \le 0$.

A unified way to write these is $\min(f(x))_{x=a}^b \le \dfrac1{b-a}\int_a^b f(x) dx \le \max(f(x))_{x=a}^b $. In words, the average is between the min and the max.

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