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I wanna prove the following identity for big values of $N\gg 1$ $$ {}_3F_1\left(-N+1,1,1;2;-\frac{1}{N}\right)\to\frac{1}{2}\bigg({}_2F_1\left(1,1;2;1-\frac{1}{N}\right)+\log 2+\gamma\bigg) $$

where $\gamma$ is the Euler-Mascheroni constant. By trying big values for $N$ in the above formula above it looks like the conjecture holds, but I am unable to prove it. Anybody can do better?

Solution attempt following Jack D'Aurizio's indications. Let's start with $$ {}_3F_1\left(-N+1,1,1;2;-\frac{1}{N}\right)=\sum_{n=0}^{\infty}\frac{(1)_n(1)_n}{(2)_n}\frac{1}{n!}(-N+1)_n\left(\frac{-1}{N}\right)^n $$ $$ =\sum_{n=0}^{\infty}\frac{1}{n+1}(-N+1)_n\left(\frac{-1}{N}\right)^n $$ $$ =\sum_{n=0}^{\infty}\frac{1}{n+1}\frac{(N-1)(N-2)\ldots(N-n)}{N^n} $$ $$ =\sum_{n=0}^{\infty}\frac{1}{n+1}\frac{\Gamma(N)}{\Gamma(N-n)}\frac{1}{N^n} $$ let's focus on $$ \frac{\Gamma(N)}{\Gamma(N-n)}\frac{1}{N^n} $$ $$ =\frac{1}{N^n}\frac{\frac{e^{-\gamma{}N}}{N}\prod_{m=1}^{\infty}\frac{m}{m+N}e^{N/m}}{\frac{e^{-\gamma{}(N-n)}}{N-n}\prod_{m=1}^{\infty}\frac{m}{m+N-n}e^{(N-n)/m}} $$ $$ =\frac{N-n}{N^{n+1}}e^{-\gamma{}n}\prod_{m=1}^{\infty}\frac{m+N-n}{m+N}e^{n/m} $$ $$ =\frac{N-n}{N^{n+1}}e^{-\gamma{}n}\prod_{m=1}^{\infty}\bigg(1-\frac{n}{m+N}\bigg)e^{n/m} $$

PhoenixPerson
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1 Answers1

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$$\begin{eqnarray*}{}_3F_1\left(-N+1,1,1;2;-\frac{1}{N}\right)&=&\sum_{n\geq 0}\frac{(-N+1)_n }{(n+1) N^n}(-1)^n\\ &=& \sum_{n\geq 0}\frac{1}{n+1}\color{red}{\left(\frac{(N-1)(N-2)\cdot\ldots\cdot(N-n)}{N^n}\right)}\end{eqnarray*}$$ while: $$ \phantom{}_2 F_1 \left(1,1;2;1-\frac{1}{N}\right) = \sum_{n\geq 0}\frac{1}{n+1}\color{red}{\left(1-\frac{1}{N}\right)^n}=\frac{\log N}{1-\frac{1}{N}} $$ so your conjecture can be proved by using the definition of the $\Gamma$ function given by Euler's product, then exploiting the definition through the Weierstrass product and applying the dominated convergence theorem.

Jack D'Aurizio
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  • Could you be a little more explicit in the end on how the Euler product is useful inbetween? – PhoenixPerson May 31 '16 at 17:08
  • You have to estimate the difference between the red terms: that is the point for the $\Gamma$ function representations come into play. – Jack D'Aurizio May 31 '16 at 17:20
  • I am sorry but I don't see how the Gamma is useful here – PhoenixPerson May 31 '16 at 17:26
  • @AnarchistBirdsWorshipFungus: $$ (a)_n = \frac{\Gamma(a+n)}{\Gamma(a)}$$ – Jack D'Aurizio May 31 '16 at 17:28
  • what do you mean with the Weirstrass product? – PhoenixPerson May 31 '16 at 17:46
  • @AnarchistBirdsWorshipFungus: what is known as the Weierstrass product, the one with $e^{-\gamma x}$ appearing as a term. Just follow the link. – Jack D'Aurizio May 31 '16 at 17:47
  • I have added where I am in the question. Further clarification would ve very appreciated – PhoenixPerson May 31 '16 at 21:41
  • How does the dominated convergence theorem help here? – PhoenixPerson Jun 01 '16 at 10:47
  • In:$\frac{1}{N^n}\frac{\frac{e^{-\gamma{}N}}{N}\prod_{m=1}^{\infty}\frac{m}{m+N}e^{N/m}}{\frac{e^{-\gamma{}(N-n)}}{N}\prod_{m=1}^{\infty}\frac{m}{m+N-n}e^{(N-n)/m}}$ you have N instead of N-n in the lower left hand corner. In the next equation you have corrected it. – rrogers Jun 01 '16 at 12:55
  • @rrogers thanks, edited. Do you know how to continue? – PhoenixPerson Jun 01 '16 at 15:46
  • @AnarchistBirdsWorshipFungus I believe that Jack is looking at something like the Alternative definitions for the Gamma functions on Wikipedia: https://en.wikipedia.org/wiki/Gamma_function#Alternative_definitions Equating them and hoping that the restriction on t not being a non-postive integer is lifted. Formally equating the two sides: $\lim_{n \to \infty} \frac{n! ; n^t}{t ; (t+1)\cdots(t+n)} = \frac{1}{t} \prod_{n=1}^\infty \frac{\left(1+\frac{1}{n}\right)^t}{1+\frac{t}{n}}= \frac{e^{-\gamma t}}{t} \prod_{n=1}^\infty \left(1 + \frac{t}{n}\right)^{-1} e^{\frac{t}{n}} $ – rrogers Jun 03 '16 at 11:37
  • @AnarchistBirdsWorshipFungus Treating on a term by term basis leads to logarithms of the exponentials. – rrogers Jun 03 '16 at 11:44
  • @rrogers could you please be more explicit on the term by term appearance of the logs? I do not follow you – PhoenixPerson Jun 03 '16 at 13:45
  • @AnarchistBirdsWorshipFungus What I meant was: $ln(1-x)=\sum\frac{1}{n}x^{n}$ and we are trying to evaluate the x^n before doing the sum/logarithm. Sorry to obscure the issue. – rrogers Jun 03 '16 at 14:08
  • @rrogers don't be sorry, any help is greatly appreciated! but I still don't know what sum you are taking about. I mean, consider the last equation in my question, how does it follow? – PhoenixPerson Jun 03 '16 at 14:16