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)=\ ?
$
 A: 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
A: Let $F(x)=\int_a^x f(t)dt$, we can obtain $\frac{dF(x)}{dx}=f(x)$ with the help of Newton-Leibniz formula. In a more easy sense, $f(x)$ is actually the height of the area. It is actually the change rate of the area.
Similarly, when $F(x)=\sum_a^x f(k)$, we can obtain that $\frac{dF(x)}{dx}=f(x)$.
A: 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
