Summation simpilification process Here is my summation:
$$\sum_{n=0}^{N/2-1}\frac1{25+nB}=A$$
where $0.01\le A\le2$, $10\le N\le2000$ and I need to find the $B$ for different values of $N$ I calculate this summation online (check here)
But the Digamma function makes the output function complex and it is not easy to solve with a calculator or even by hand. I need to calculate the result of this summation with hand easily. Is there any way to convert my summation to simple formula? Something like this:
\begin{align*}
\sum^n_{k=1}k&=\frac{n(n+1)}{2}\\[5pt]
\sum^n_{k=1}k^2&=\frac{n(n+1)(2n+1)}{6}\\[5pt]
\end{align*}
 A: Exact Solution:
Consider the function:
$$
I'(x) = \sum_{k=0}^n\frac{1}{x+kb}
$$
If we integrate both sides:
$$
I(x) + C = \sum_{k=0}^n\log\left(x+kb\right) = \log\prod_{k=0}^n\left(x+kb\right) = \log\frac{(x+nb)!}{(x-1)!}
$$
In other words:
$$
I(x) + C = \log\Gamma(x+nb+1) - \log\Gamma(x)
$$
If we take the derivative on both sides once more:
$$
I'(x) = \psi(x+nb+1) - \psi(x),\quad\text{where}\quad
\psi(x) = \frac{d}{dx}\log\Gamma(x)
$$
Where $\psi(x)$ is known as digamma function. And that's why your summation depends on the digamma function, because its the derivative of $I(x)$ at $x=25$. Thus, there's no way to convert the sum to a simple formula, as that would be equivalent of saying there exists a simple formula for the digamma function. However, there's the possibility of using an approximation.

Approximate Solution:
Asymptotic expansion of the gamma function can be written like this:
$$\ln\Gamma(x) = x\ln x - x + \frac{1}{2}\log\frac{2\pi}{x} + O\left(\frac{1}{x}\right)$$
$$\psi(x) = \ln x - \frac{1}{2x} + O\left(\frac{1}{x^2}\right) $$
Thus, an approximate solution is:
$$
I'(x) = \ln (x+nb+1) - \ln (x) + \frac{1}{2x} - \frac{1}{2(x+nb+1)} + O\left(\frac{1}{x^2}\right)
$$
$$
I'(x) = \ln\left(1 + \frac{nb+1}{x}\right) + \frac{nb+1}{2x(x+1+nb)} +  O\left(\frac{1}{x^2}\right)
$$
$$
I'(x) = \ln\left(1 + \frac{nb+1}{x}\right) + O\left(\frac{1}{x^2}\right)
$$
=]

To translate this to your situation, let $A = I'(25)$, $b=B$, $x=25$ and $n = N/2-1$, and thus you get:
$$
A \approx \ln\left(1 + \frac{1-B}{25} + \frac{NB}{50}\right)
$$
=]
A: Your sum is
$$
A = \psi \left( {\left( {\tfrac{N}{2} - 1} \right)B + 26} \right) - \psi (25).
$$
It is not possible to express your sum in terms of elementary functions. Nevertheless, one can obtain good upper and lower bounds for $B$. It is known that
$$
\log \left( {x + \tfrac{1}{2}} \right) < \psi (x + 1) < \log (x+1)
$$
for all $x>0$. Thus
$$
\log \left( {\left( {\tfrac{N}{2} - 1} \right)B + \tfrac{{51}}{2}} \right) < A + \psi (25) < \log \left( {\left( {\tfrac{N}{2} - 1} \right)B + 26} \right),
$$
i.e.,
$$
\frac{{2e^{A + \psi (25)}  - 52}}{{N - 2}} < B < \frac{{2e^{A + \psi (25)}  - 51}}{{N - 2}},
$$
with $\psi(25)=3.198742512\ldots$. For example, with $A=1$ and $N=500$, we get
$$
0.163 \ldots  < B < 0.165 \ldots .
$$
You can go beyond this, by using asymptotic approximations for the inverse of the digamma function (cf. https://doi.org/10.1007/s11139-014-9659-3). In particular,
$$
B \approx \frac{{2e^{A + \psi (25)}  - 51}}{{(N - 2)}} - \frac{1}{{12(N - 2)e^{A + \psi (25)} }}.
$$
With $A=1$ and $N=500$, this gives $B=0.16506787525\ldots$, whereas the true value is $B = 0.16506787531\ldots$.
