Finding $2 \pi \sum_{i=0}^N \left(1 - \frac{a}{a + b - i c}\right)^{\frac{3}{2}}$

Asked Apr 29 '21 at 18:29

Active Apr 29 '21 at 19:54

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Is it possible to solve the following series?

$$2 \pi \sum_{i=0}^N \left(1 - \frac{a}{a + b - i c}\right)^{\frac{3}{2}}$$

where $a,b,c$ are parameters.

Since $a \gg b,c,N$, solving the Taylor expansion might also do the trick:

$$3 \pi \sum_{i=0}^N \left(1 - \frac{a}{a + b - i c}\right)$$

But I'm not able to solve this either.

I don't think it helps but the upper limit is:

$N = \frac{b}{c}$

edited Apr 29 '21 at 19:54

StubbornAtom

asked Apr 29 '21 at 18:29

Babababa31

2

I don’t see any reason to think there is a closed form. You might be able to get upper and lower bounds. – Thomas Andrews Apr 29 '21 at 18:34
Wolfram gives this solution but it's rather complicate – Babababa31 Apr 29 '21 at 18:38
And I can't see the step-by-step solution – Babababa31 Apr 29 '21 at 18:40
Here a correct Wolfram Alpha answer. Even more complicated. And this is without the $3/2$ exponent. Link – Thomas Andrews Apr 29 '21 at 18:54
1

The digamma function Wofram Alpha uses in the answer is a frequent way to hide a finite sum. For example, for integers $n$:$$\psi^{(0)}(n)=\psi(n)=\left(\sum_1^{n-1}\frac1k\right)-\lambda$$ https://en.wikipedia.org/wiki/Digamma_function – Thomas Andrews Apr 29 '21 at 19:05
Yeah I guess there is no solution apart from that one, thanks anyway – Babababa31 Apr 29 '21 at 19:21
1

In particular, the WA answer has the term: $\psi(x+y)-\psi(x)$ and: $$\psi(x+y)-\psi(x)=\frac1{x+1}+\cdots+\frac1{ x+y-1}$$ – Thomas Andrews Apr 29 '21 at 19:23

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