Prove it.

Calculus Level 5

$\large\sqrt[4]{2}\times\sqrt[9]{3}\times\sqrt[16]{4}\times\sqrt[25]{5}\times\cdots={\left(\frac{A^P}{2\pi e^\gamma}\right)}^{\frac{\pi^2}{Q}}$

The equation above holds true for positive integers $P$ and $Q$ . Submit your answer as $P+Q$ .

Notations:

$A$ denotes the Glaisher–Kinkelin constant .
$\gamma$ denotes the Euler–Mascheroni constant .

The answer is 18.

1 solution

Mark Hennings
Nov 20, 2017

The logarithm of the product is $\sum_{n=2}^\infty \frac{\ln n}{n^2} \; = \; - \zeta'(2) \; = \; \tfrac{1}{6}\pi^2\big(12\ln A - \ln 2\pi - \gamma\big)$ which makes the product equal to $\left(\frac{A^{12}}{2\pi e^\gamma}\right)^{\frac16\pi^2}$ so the answer is $12+6=\boxed{18}$ .

Why is $\; - \zeta'(2) \; = \; \tfrac{1}{6}\pi^2\big(12\ln A - \ln 2\pi - \gamma\big)$ ?

Digvijay Singh - 3 years, 6 months ago

That goes back, essentially,. to the original 1878 paper by Glaisher which introduced the constant.

Mark Hennings - 3 years, 6 months ago

Prove it.

The answer is 18.

1 solution

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