Editing Euler's constant (section)

=== Exponential ===
The constant {{math|{{var|e}}{{i sup|1px|{{var|γ}}}}}} is important in number theory.  Its numerical value is:{{r|OEIS_A073004}}

{{block indent|{{gaps|1.78107|24179|90197|98523|65041|03107|17954|91696|45214|30343|...}}.}}
{{math|{{var|e}}{{i sup|1px|{{var|γ}}}}}} equals the following [[limit of a sequence|limit]], where {{math|{{var|p}}{{sub|{{var|n}}}}}} is the {{mvar|n}}th [[prime number]]:

<math display="block">e^\gamma = \lim_{n\to\infty}\frac1{\log p_n} \prod_{i=1}^n \frac{p_i}{p_i-1}.</math>

This restates the third of [[Mertens' theorems]].{{r|excursions}}

We further have the following product involving the three constants {{math|{{var|e}}}}, {{math|{{var|π}}}} and {{math|{{var|γ}}}}:<ref name=":9">{{Cite web |last=Weisstein |first=Eric W. |title=Mertens Theorem |url=https://mathworld.wolfram.com/MertensTheorem.html |access-date=2024-10-08 |website=mathworld.wolfram.com |language=en}}</ref>

<math display="block">\frac{\pi^2}{6e^\gamma}=\lim_{n\to\infty}\frac1{\log p_n} \prod_{i=1}^n \frac{p_i}{p_i+1}.</math>

Other [[infinite product]]s relating to {{math|{{var|e}}{{i sup|1px|{{var|γ}}}}}} include:

<math display="block">\begin{align}
\frac{e^{1+\frac{\gamma}{2}}}{\sqrt{2\pi}} &= \prod_{n=1}^\infty e^{-1+\frac1{2n}}\left(1+\frac1{n}\right)^n \\
\frac{e^{3+2\gamma}}{2\pi} &= \prod_{n=1}^\infty e^{-2+\frac2{n}}\left(1+\frac2{n}\right)^n. \end{align}</math>

These products result from the [[Barnes G-function|Barnes {{mvar|G}}-function]].

In addition,

<math display="block">e^{\gamma} = \sqrt{\frac2{1}} \cdot \sqrt[3]{\frac{2^2}{1\cdot 3}} \cdot \sqrt[4]{\frac{2^3\cdot 4}{1\cdot 3^3}} \cdot \sqrt[5]{\frac{2^4\cdot 4^4}{1\cdot 3^6\cdot 5}} \cdots</math>

where the {{mvar|n}}th factor is the {{math|({{var|n}} + 1)}}th root of

<math display="block">\prod_{k=0}^n (k+1)^{(-1)^{k+1}{n \choose k}}.</math>

This infinite product, first discovered by Ser in 1926, was rediscovered by Sondow using [[hypergeometric function]]s.{{r|Sondow2003}}

It also holds that{{r|ChoiSrivastava2010}}

<math display="block">\frac{e^\frac{\pi}{2}+e^{-\frac{\pi}{2}}}{\pi e^\gamma}=\prod_{n=1}^\infty\left(e^{-\frac{1}{n}}\left(1+\frac{1}{n}+\frac{1}{2n^2}\right)\right).</math>