For the moment, consider the corresponding problem involving integration. Let $s(x)$ be the explicit solution to the following integral.
$
\displaystyle s(x)=\int_a^x f(t) \, dt
$
The function $s'(x)$ is equivalent to the derivative of the integral with respect to it's upper limit and may be expressed in integral form.
$
\displaystyle s'(x)=\partial _x\left(\int_a^x f(t) \, dt\right)=f(a)+\int_a^x f'(t) \, dt
$
Now let $s(x)$ be the explicit solution to the following summation.
$
\displaystyle s(x)=\sum _{t=a}^x f(t)
$
The function $s'(x)$ is equivalent to the derivative of the summation with respect to it's upper limit. What is the derivative of $s(x)$ expressed in summation form?
$
\displaystyle s'(x)=\partial _x\left(\sum _{t=a}^x f(t)\right)=\ ?
$
Best Answer
In order to express continuous summation, define an auxiliary function $s(x,f)$ equal to the $RHS$ of the Euler-Maclaurin formula with zero remainder, such that, the index variable $n$ is replaced with the real variable $x$. $$ \displaystyle s(x,f)\text{:=}\int_a^x f(t) \, dt+\frac{f(x)}{2}+\frac{f(a)}{2}+\sum _{k=1}^{\infty } \frac{B_{2 k} }{(2 k)!} \left(f^{(2 k-1)}(x)-f^{(2 k-1)}(a)\right) $$
Clearly, if $x$ is an integer $n$, then $s(n,f)$ is equal to the summation of $f(t)$.
$ \displaystyle s(n,f)=\sum _{t=a}^n f(t) $
We can now derive a differential equation for $s$. Differentiate $s$ with respect to $x$ and simplify.
$ \displaystyle \frac{\partial s(x,f)}{\partial x}=f(x)+\frac{f'(x)}{2}+\sum _{k=1}^{\infty } \frac{B_{2 k} f^{(2 k)}(x)}{(2 k)!} $
Substitute $f'(t)$ for $f(t)$.
$ \displaystyle s\left(x,f'\right)=f(x)-f(a)+\frac{f'(x)}{2}+\frac{f'(a)}{2}+\sum _{k=1}^{\infty } \frac{B_{2 k} }{(2 k)!} \left(f^{(2 k)}(x)-f^{(2 k)}(a)\right) $
Calculate the difference and simplify.
$ \displaystyle \frac{\partial s(x,f)}{\partial x}-s\left(x,f'\right)=\sum _{k=0}^{\infty } \frac{B_k}{k!} f^{(k)}(a) $
Observe that the term $\frac{f^{(k)}(a)}{k!}$ is the coefficient for the Taylor expansion of $f(t)$, hence.
$ \displaystyle f(t)=\sum _{k=0}^{\infty } \frac{f^{(k)}(a)}{k!} (t-a)^k=\sum _{k=0}^{\infty } c(k) (t-a)^k $
$ \displaystyle \frac{\partial s(x,f)}{\partial x}-s\left(x,f'\right)=\sum _{k=0}^{\infty } c(k) B_k $
Consider what happens if we restrict $x$ to be an integer $n$.
$ \displaystyle \left.\frac{\partial s(x,f)}{\partial x}\right|_{x=n}=\sum _{k=0}^{\infty } c(k) B_k+\sum _{t=a}^n f'(t) $
Now, let's get creative and allow the following symbolic equivalence.
$ \displaystyle \left.\frac{\partial s(x,f(t))}{\partial x}\right|_{x=n}\equiv \partial _n\sum _{t=a}^n f(t) $
The derivative of a summation with respect to it's upper limit may then be expressed as,
$$ \displaystyle \partial _n\sum _{t=a}^n f(t)=\sum _{k=0}^{\infty } c(k) B_k+\sum _{t=a}^n f'(t) $$
Example
As an example consider the following summation identity with integer $m\geq 1$.
$ \displaystyle \sum _{t=0}^n t^m=\frac{B_{m+1}(n+1)-B_{m+1}}{m+1} $
Define the functions $f$ and $s$.
$ \displaystyle f(t)=t^m $
$ \displaystyle s(x,f)=\frac{B_{m+1}(x+1)-B_{m+1}}{m+1} $
Verify the differential equation.
$ \displaystyle \partial _n\sum _{t=a}^n f(t)=\sum _{k=0}^{\infty } c(k) B_k+\sum _{t=a}^n f'(t) $
$ \displaystyle \left.\partial _x\left(\frac{B_{m+1}(x+1)-B_{m+1}}{m+1}\right)\right|_{x=n}=B_m+\sum _{t=0}^n m t^{m-1} $
$ \displaystyle B_m(n+1)=B_m+m\left(\frac{(B_m(n+1)-B_m)}{m}\right) $
$ \displaystyle B_m(n+1)=B_m(n+1) $
Application
As an application consider the following summation identity that uses the incomplete gamma function, $\Gamma (a,z)$.
$ \displaystyle \sum _{t=0}^n \frac{1}{\Gamma (t+1)}=\frac{e \Gamma (n+1,1)}{n!} $
Define the functions $f$ and $s$.
$ \displaystyle f(t)=\frac{1}{\Gamma (t+1)} $
$ \displaystyle s(x,f)=\frac{e \Gamma (x+1,1)}{\Gamma (x+1)} $
Let the constant $C$ represent the Bernoulli sum.
$ \displaystyle C=\sum _{k=0}^{\infty } c(k) B_k $
Because $C$ is a constant, we may choose any valid $n$ to solve for $C$, choose $n=0$.
$ \displaystyle C=-e \text{Ei}(-1) $
In conclusion, assemble the differential equation and simplify. The variable $t$ is replaced with the standard index variable $k$. The result is a summation identity that makes use of these additional functions: (1) $\, _2\tilde{F}_2$, (2) $\text{Ei}$, and (3) $\psi ^{(0)}$.
$$ \displaystyle \frac{-1}{e}\sum _{k=0}^n \frac{1}{k!}\psi ^{(0)}(k+1)=\text{Ei}(-1)+\psi ^{(0)}(n+1) \left(1-\frac{1}{n!}\Gamma (n+1,1)\right)\\ +n! \, _2\tilde{F}_2(n+1,n+1;n+2,n+2;-1) $$
Peace