New answers tagged hypergeometric-functions
3
votes
How did Ramanujan find $\sum_{n=0}^\infty (-1)^n\frac{(1/2)_n(1/4)_n(3/4)_n}{n!^3}\frac{644n+41}{25920^n}=\frac{288\sqrt{5}}{5\pi}?$
The formula in the OP is a hypergeometric function identity,
$$\frac{1}{41}\sum_{n=0}^\infty (-1)^n\frac{(1/2)_n(1/4)_n(3/4)_n}{n!^3}\frac{644n+41}{25920^n}=$$
$$\qquad= \,_4F_3\left(\frac{1}{2},\frac{...
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6
votes
Accepted
How did Ramanujan find $\sum_{n=0}^\infty (-1)^n\frac{(1/2)_n(1/4)_n(3/4)_n}{n!^3}\frac{644n+41}{25920^n}=\frac{288\sqrt{5}}{5\pi}?$
There are (at least) two ways to prove Ramanujan-like formulas for
$1/\pi$: either using elliptic functions and modular equations, as
you try above, or by using modular functions, which is not the ...
- 11.7k
1
vote
Accepted
Factorial series $j(D)=\sum_{n=1}^\infty \frac{1}{(n^D)!}$ and hypergeometric functions
The bug is in Maxima's simplify_sum - which is used by SageMath
...
- 13.7k
3
votes
Accepted
Ask for a generating function or an explicit expression of a triangle of positive integers
The generating function:
$${\cal C}(x,y) = \sum_{n,k\geq 0} C_{n,k} x^n y^{2k}$$
has the following explicit form:
$${\cal C}(x,y) = \frac{\arctan(y)}{y(1-x(1+y^2))}.$$
For "one more problem",...
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