What is the derivative of a summation with respect to its upper limit? 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)=\ ?
$
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
Consider the following integral equation with constant $C$.

$
\displaystyle \partial _x\left(\int_a^x f(t) \, dt\right)-\int_a^x f'(t) \, dt=C
$

Is there a corresponding summation equation?

$
\displaystyle \partial _x\left(\sum _{t=a}^x f(t)\right)-\sum _{t=a}^x f'(t)=C
$

The proof of this result is based on the Euler-Maclaurin formula assuming zero remainder.

$
\displaystyle \sum _{t=a}^x f(t)=\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)
$

Differentiate with respect to x.

$
\displaystyle \partial _x\left(\sum _{t=a}^x f(t)\right)=f(x)+\frac{f'(x)}{2}+\sum _{k=1}^{\infty } \frac{B_{2 k}}{(2 k)!} f^{(2 k)}(x)
$

Apply the formula to $f'(t)$.

$
\displaystyle \sum _{t=a}^x f'(t)=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 \partial _x\left(\sum _{t=a}^x f(t)\right)-\sum _{t=a}^x f'(t)=\sum _{k=0}^{\infty } \frac{B_k}{k!} f^{(k)}(a)\text{  }\ \square
$

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
$

The derivative of a summation with respect to it's upper limit is therefore,
$$
\displaystyle \partial _x\left(\sum _{t=a}^x f(t)\right)=\sum _{k=0}^{\infty } c(k)B_k+\sum _{t=a}^x f'(t)
$$

Example

As an example let $f(t)=e^{r t}$ and $a=0$. The summation of $e^{r t}$ is just the geometric series in $e^{r}$, hence.

$
\displaystyle \sum _{t=0}^x f(t)=\sum _{t=0}^x e^{r t}=\sum _{t=0}^x \left(e^r\right)^t=\frac{\left(e^r\right)^{x+1}-1}{e^r-1}=\frac{e^{r (x+1)}-1}{e^r-1}
$

$
\displaystyle \partial _x\left(\sum _{t=0}^x f(t)\right)=\partial _x\left(\sum _{t=0}^x e^{r t}\right)=\partial _x\left(\frac{e^{r (x+1)}-1}{e^r-1}\right)=\frac{r e^{r (x+1)}}{e^r-1}
$

$
\displaystyle \sum _{t=0}^x f'(t)=\sum _{t=0}^x r e^{r t}=\frac{r\left(e^{r (x+1)}-1\right)}{e^r-1}
$

The Bernoulli sum in terms of the coefficient function $c(k)$.

$
\displaystyle e^{r t}=\sum _{k=0}^{\infty } \frac{r^k}{k!} t^k
$

$
\displaystyle c(k)=\frac{r^k}{k!}
$

$
\displaystyle \sum _{k=0}^{\infty } c(k) B_k=\sum _{k=0}^{\infty } \frac{r^k}{k!} B_k
$

Notice that this is the generating form for the Bernoulli numbers.

$
\displaystyle \frac{r}{e^r-1}=\sum _{k=0}^{\infty } \frac{B_k}{k!} r^k
$

Now assemble the equation and simplify.

$
\displaystyle \partial _x\left(\sum _{t=a}^x f(t)\right)=\sum _{k=0}^{\infty } c(k)B_k+\sum _{t=a}^x f'(t)
$

$
\displaystyle \frac{r e^{r (x+1)}}{e^r-1}=\frac{r}{e^r-1}+\frac{r\left(e^{r (x+1)}-1\right)}{e^r-1}
$

$
\displaystyle r=r
$

Peace