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*}
Best Answer
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: $$ A \approx \ln\left(1 + \frac{1-B}{25} + \frac{NB}{50}\right) $$
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