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In Mathworld's article Gamma function, in line (96), we find the formula,

$\sum_{k=0}^\infty (8k+1)\left(\frac{\Gamma(k+\frac{1}{4})}{k!\;\Gamma(\frac{1}{4})}\right)^4 = 2^{3/2}\frac{1}{\sqrt{\pi}\,\left(\Gamma(3/4)\right)^2}$

On a whim, I evaluated the LHS and RHS using Mathematica to 100-digit precision, and found the first few digits as,

$\text{LHS} = 1.062679901\dots$

$\text{RHS} = 1.062679899\dots$

Ahem, they don't match. If it is a typo, then I find it interesting it is exceedingly close.

So what is the problem? 1) Did I input it in Mathematica wrongly? 2) Is there a typo, or misplaced symbol by authors after Ramanujan (Weisstein gives Hardy et al as references) 3) Or was Ramanujan just mistaken?

2 Answers 2

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The series in question converges slowly, $(8k+1) \left( \frac{(1/4)_k}{k!} \right)^4 \sim \frac{8}{k^2 \Gamma^4(1/4)}$, hence it may be that you have not computed enough terms.

The sum represents a value of a hypergeometric function: $ \sum_{k=0}^\infty (8k+1) \left( \frac{(1/4)_k}{k!} \right)^4 = {}_4F_3\left( \frac{1}{4},\frac{1}{4}, \frac{1}{4}, \frac{1}{4}; 1,1,1 | 1\right) - \frac{1}{32} {}_4F_3\left( \frac{5}{4},\frac{5}{4}, \frac{5}{4}, \frac{5}{4}; 2,2,2 | 1\right) $ Evaluating these numerically agrees with the expression in terms of $\Gamma$ constant:

In[18]:= Sum[(8 k + 1) (Pochhammer[1/4, k]/k!)^4, {k, 0, \[Infinity]}]  Out[18]= 1/32 (32 HypergeometricPFQ[{1/4, 1/4, 1/4, 1/4}, {1, 1, 1},       1] + HypergeometricPFQ[{5/4, 5/4, 5/4, 5/4}, {2, 2, 2}, 1])  In[19]:= N[%, 30]  Out[19]= 1.06267989991684365118249019510  In[20]:= N[(2^(3/2)/(Sqrt[Pi] Gamma[3/4]^2)), 30]  Out[20]= 1.06267989991684365118249019510 
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    Good God, and here I thought it was infallible. I better buy a new one then. (If I can scrape up the dough.) :-(2012-05-07
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FWIW, Maple 16 gets the sum right (symbolically, and numerically to within a reasonable roundoff error).

L:= Sum((8*k+1)*(GAMMA(k+1/4)/k!/GAMMA(1/4))^4,k=0..infinity);

R:= value(L);

$ R := 2\,{\frac {\sqrt {2}}{\sqrt {\pi } \left( \Gamma \left( 3/4 \right) \right) ^{2}}}$

evalf(L - R, 100);

$ 0.1\ 10^{-98} $

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    Thanks, Dr. Israel. I was using an old Mathematica (Version 4) and unfortunately, even with 100-digit precision, it gave me an erroneous value. Close, but not quite.2012-05-07