Let $a$ and $b$ be integers such that $a<b$ , and let $f:[a,b]\to\mathbb{R}$ be continuous. We will prove by induction that for all integers $k\ge 0$ , if $f$ is a $C^{k+1}$ function, \begin{equation} \label{em1} \sum_{a<n\le b}f(n) = \int_a^b f(t)\dd t + \sum_{r=0}^k \frac{(-1)^{r+1}B_{r+1}}{(r+1)!} (f^{(r)}(b)-f^{(r)}(a)) + \frac{(-1)^k}{(k+1)!} \int_a^b B_{k+1}(t)f^{(k+1)}(t)\dd t \end{equation}where $B_r$ is the $r$ th Bernoulli number and $B_r(t)$ is the $r$ th Bernoulli periodic function.

To prove the formula for $k=0$ , we first rewrite $\int_{n-1}^{n}f(t)\dd t$ , where $n$ is an integer, using integration by parts: \begin{eqnarray*} \int_{n-1}^n f(t)\dd t&=&\int_{n-1}^n\frac{\dd}{\dd t} (t-n+\frac{1}{2})f(t)\dd t\\ &=&(t-n+\frac{1}{2})f(t)\big\vert_{n-1}^n - \int_{n-1}^n (t-n+\frac{1}{2})f'(t)\dd t\\ &=&\frac{1}{2}(f(n)+f(n-1))-\int_{n-1}^n (t-n+\frac{1}{2})f'(t)\dd t. \end{eqnarray*}Because $t-n+\frac{1}{2}=B_1(t)$ on the interval $(n-1,n)$ , this is equal to $$ \int_{n-1}^n f(t)\dd t=\frac{1}{2}(f(n)+f(n-1))-\int_{n-1}^n B_1(t)f'(t)\dd t. $$ From this, we get $$ f(n)=\int_{n-1}^n f(t)\dd t+\frac{1}{2}(f(n)-f(n-1))+ \int_{n-1}^n B_1(t)f'(t)\dd t. $$ Now we take the sum of this expression for $n=a+1,a+2,\ldots,b$ , so that the middle term on the right telescopes away for the most part: $$ \sum_{n=a+1}^b f(n)=\int_a^b f(t)\dd t+\frac{1}{2}(f(b)-f(a))+ \int_a^b B_1(t)f'(t)\dd t $$ which is the Euler-Maclaurin formula for $k=0$ , since $B_1=-\frac{1}{2}$ .

Suppose that $k>0$ and the formula is correct for $k-1$ , that is \begin{equation} \label{em2} \sum_{a<n\le b}f(n) = \int_a^b f(t)\dd t + \sum_{r=0}^{k-1} \frac{(-1)^{r+1} B_{r+1}}{(r+1)!} (f^{(r)}(b)-f^{(r)}(a)) + \frac{(-1)^{k-1}}{k!} \int_a^b B_k(t)f^{(k)}(t)\dd t. \end{equation}We rewrite the last integral using integration by parts and the facts that $B_k$ is continuous for $k\ge 2$ and $B_{k+1}'(t)=(k+1)B_k(t)$ for $k\ge 0$ : \begin{eqnarray*} \int_a^b B_k(t)f^{(k)}(t)\dd t&=&\int_a^b \frac{B_{k+1}'(t)}{k+1} f^{(k)}(t) \dd t\\ &=&\frac{1}{k+1}B_{k+1}(t)f^{(k)}(t)\big\vert_a^b - \frac{1}{k+1} \int_a^b B_{k+1}(t) f^{(k+1)}(t)\dd t. \end{eqnarray*}Using the fact that $B_k(n)=B_k$ for every integer $n$ if $k\ge 2$ , we see that the last term in Eq. is equal to $$ \frac{(-1)^{k+1}B_{k+1}}{(k+1)!}(f^{(k)}(b)-f^{(k)}(a)) + \frac{(-1)^k}{(k+1)!} \int_a^b B_{k+1}(t) f^{(k+1)}(t)\dd t. $$ Substituting this and absorbing the left term into the summation yields Eq. , as required.