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Hi could you demostrated that $$\sum _{m=1}^{\infty } (-1)^{m+1} \log (\Gamma (m+1))=\frac{1}{4} \log \left(\frac{2}{\pi }\right)$$ and $$\sum _{m=1}^{\infty } \log (\Gamma (m+1))=\log \left((2 \pi )^{3/4} A\right)-\frac{3}{4}$$ I know this series are divergent.

capea
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    If they are divergent, then what do the equations mean? and what is $A$? – Gerry Myerson Sep 16 '13 at 10:51
  • Hi A is glassier constant i think ist possible check numerecally using euler.boole sum formula. – capea Sep 16 '13 at 10:56
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    Do you mean the Glaisher constant, http://en.wikipedia.org/wiki/Glaisher–Kinkelin_constant ? I'm not sure how you check a divergent series numerically, but then I don't know the Euler-Boole sum formula. – Gerry Myerson Sep 16 '13 at 10:59
  • I am hesitant to downvote the question. @capea, can you clarify what you mean by "equality" if you know both series are divergent? Because if you know they are divergent, writing an equality up there doesn't work in the classical sense. – Patrick Da Silva Sep 16 '13 at 11:13
  • HI you can see http://www.jstor.org/discover/10.2307/40391116?uid=3737952&uid=2&uid=4&sid=21102653915233 – capea Sep 16 '13 at 13:58
  • @capea You need to pay to see the article about your question, so please explain clearly your problem so everyone can understand. – user37238 Sep 17 '13 at 08:36
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    I think the first sum is a duplicate of http://math.stackexchange.com/questions/272083/kind-of-counter-intuitive-sum-of-log-gamma – Ron Gordon Sep 17 '13 at 10:00

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The sum can have the following integral representation

$$\sum _{m=1}^{\infty } (-1)^{m+1} \log (\Gamma (m+1))=-\frac{1}{4}\,\int _{0}^{1}\!{\frac {u\ln \left( u \right)+\ln \left( u \right) -2\,u+2}{\ln \left( u \right) \left( {u}^{2}-1 \right) }}{d u}-\frac{\gamma}{4}\sim -.1128956763. $$

where $\gamma$ is the Euler constant.

Mhenni Benghorbal
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  • In what sense are those three expressions equal? If that integral converges, how does it represent the divergent sum? If the integral doesn't converge, how does it equal that decimal? – Gerry Myerson Sep 17 '13 at 13:09