Euler-Maclaurin Summation Formula

Simplified Version

Theorem

Let $f$ be a real function which is continuous, positive and decreasing on the interval $\left[{1 .. \infty}\right)$.

Let the sequence $\left \langle {\Delta_n} \right \rangle$ be defined as:

$\displaystyle \Delta_n = \sum_{k=1}^n f \left({k}\right) - \int_1^n f \left({x}\right) \ \mathrm dx$

Then $\left \langle {\Delta_n} \right \rangle$ is decreasing and bounded below by zero.

Hence it converges.

Proof

From Upper and Lower Bounds of Integral, we have that:

$\displaystyle m \left({b - a}\right) \le \int_a^b f \left({x}\right) \ \mathrm dx \le M \left({b - a}\right)$

where:

• $M$ is the maximum and
• $m$ is the minimum

of $f \left({x}\right)$ on $\left[{a .. b}\right]$.

Since $f$ decreases, $M = f \left({a}\right)$ and $m = f \left({b}\right)$.

Thus it follows that:

$\displaystyle \forall k \in \N^*: f \left({k+1}\right) \le \int_k^{k+1} f \left({x}\right) \ \mathrm dx \le f \left({k}\right)$

as $\left({k+1}\right) - k = 1$.

Thus:

	\(\displaystyle \)	\(\displaystyle \Delta_{n+1} - \Delta_n\)	\(=\)	\(\displaystyle \left({\sum_{k=1}^{n+1} f \left({k}\right) - \int_1^{n+1} f \left({x}\right) \ \mathrm dx}\right) - \left({\sum_{k=1}^n f \left({k}\right) - \int_1^n f \left({x}\right) \ \mathrm dx}\right)\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle f \left({n+1}\right) - \int_n^{n+1} f \left({x}\right) \ \mathrm dx\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\le\)	\(\displaystyle f \left({n+1}\right) - f \left({n+1}\right)\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 0\)	\(\displaystyle \)

Thus $\left \langle {\Delta_n} \right \rangle$ is decreasing.

$\Box$

Also:

	\(\displaystyle \)	\(\displaystyle \Delta_n\)	\(=\)	\(\displaystyle \sum_{k=1}^n f \left({k}\right) - \sum_{k=1}^{n-1} \int_k^{k+1} f \left({x}\right) \ \mathrm dx\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\ge\)	\(\displaystyle \sum_{k=1}^n f \left({k}\right) - \sum_{k=1}^{n-1} f \left({k}\right)\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle f \left({n}\right)\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\ge\)	\(\displaystyle 0\)	\(\displaystyle \)

$\Box$

Hence the result.

$\blacksquare$

Notes

It follows from this that if $f$ is continuous, positive and decreasing on $\left[{1 .. \infty}\right)$, then the series $\displaystyle \sum_{k=1}^\infty f \left({k}\right)$ and the improper integral $\displaystyle \int_1^{\to +\infty} f \left({x}\right) \ \mathrm dx$ either both converge or both diverge.

So this theorem provides a test for the convergence of both a series and an improper integral.

For this reason, this is also sometimes called the integral test.

Source of Name

This entry was named for Leonhard Paul Euler and Colin Maclaurin.

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 13.32$