# Proof of Euler-Maclaurin summation formula

Let $a$ and $b$ be integers such that $a, and let $f:[a,b]\to\mathbf{R}$ be continuous. We will prove by induction that for all integers $k\geq 0$, if $f$ is a $k+1$ times continuously differentiable function,

$\displaystyle \sum_{a

where $B_i$ is the $i$-the Bernoulli number and $B_i(t)$ is the $i$-th Bernoulli periodic function.

To prove the formula  for $k=0$, we first write $\int_{n-1}^n f(t)\,dt$, where $n$ is an integer, using integration by parts:

$\displaystyle \int_{n-1}^n f(t)\,dt=\int_{n-1}^n (t-n+\frac{1}{2})'f(t)\,dt=\frac{1}{2}(f(n)-f(n-1))-\int_{n-1}^n (t-n+\frac12)f'(t)\,dt.$

Because $t-n+\frac12=B_1(t)$ on the interval $(n-1,n)$, this is equal to

$\displaystyle \int_{n-1}^n f(t)\,dt=\frac{1}{2}(f(n)-f(n-1))-\int_{n-1}^n B_1(t)f'(t)\,dt.$