Multiply

Calculus Level 5

$\Large \dfrac{2^{1/4}\cdot4^{1/16} \cdot 6^{1/36} \cdot 8^{1/64} \cdots}{\sqrt[3]{1^{1/1} \cdot 3^{1/9} \cdot 5^{1/25}\cdot 7^{1/49} \cdots}} = ?$

\sqrt[3]{\pi}

e^{1/\pi}

\dfrac{4\pi^2}{27}

\dfrac{2e}{1+e}

2^{\pi^2/18}

1 solution

Mark Hennings
May 2, 2018

Note that $\begin{aligned} S & = \; \sum_{n=1}^\infty \frac{\ln n}{n^2} \; = \; -\zeta'(2) \\ E & = \; \sum_{n=1}^\infty \frac{\ln (2n)}{(2n)^2} \; = \; \tfrac14\left[\ln2\,\zeta(2) + S\right] \; = \; \tfrac{1}{24}\pi^2\ln2 - \tfrac14\zeta'(2) \\ O & = \; \sum_{n=1}^\infty \frac{\ln(2n-1)}{(2n-1)^2} \; = \; S - E \; = \; -\tfrac{1}{24}\pi^2\ln2 + \tfrac34\zeta'(2) \\ E - \tfrac13O & = \; \tfrac{1}{18}\pi^2 \ln2 \end{aligned}$ and so the result we want is $e^{E - \frac13O} \; = \; \boxed{2^{\frac{1}{18}\pi^2}}$

How is $O=S-E$ ?

Digvijay Singh - 3 years, 1 month ago

$S$ is the sum of $\ln n/n$ for all positive integers $n$ , while $E$ is the sum for all even positive integers, and $O$ is the sum for all odd positive integers. Thus $S = E + O$ .

Mark Hennings - 3 years, 1 month ago

S= -f(2)

What is that function on left of 2?? (Sorry I don't have it in my keyboard so wrote f)

Mr. India - 2 years, 2 months ago

$\zeta(s) = \sum_{r=1}^\infty r^{-s}$ is the Zeta function. I state that $S = -\zeta'(2)$ is minus the derivative of $\zeta$ at $2$ .

I don't have that symbol on my keyboard, either but typing \zeta within LaTeX delimiters (consult the fomatting guide) does the business.

Mark Hennings - 2 years, 2 months ago

Thank you $\zeta{sir}$ !

Mr. India - 2 years, 2 months ago

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