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Sum of \(\frac{\ln(n)}{n^2}\)

\[\sum_{n=1}^{\infty} \dfrac{\ln(n)}{n^2} = \dfrac{\pi^2}{6}\left(12\ln(A) - \gamma - \ln(2\pi) \right)\]

Just for fun, I entered the above sum into WolframAlpha, and that's the answer it gave me. But I'm curious how we can actually prove that the sum has this closed form. Any ideas or solutions would be appreciated.

Clarifications:

\(A\) is the Glaisher-Kinkelin constant: \[A = \lim_{n\to\infty} \dfrac{e^{n^2/4}}{n^{n^2/2+n/2+1/12}}\prod_{k=1}^{n} k^k\]

\(\gamma\) is the Euler-Mascheroni constant: \[\gamma = \sum_{n=1}^{\infty} \left(\dfrac{1}{n} - \ln\left(1 + \dfrac{1}{n} \right) \right)\]

Note by Ariel Gershon
1 year, 2 months ago

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The given sum also equals \(-\zeta^{'}(2)\)

In general, \[\sum_{n=1}^{\infty} \dfrac{\ln^{a}(n)}{n^b}=(-1)^{a}\zeta^{(a)}(b)\]

where \(\zeta^{(x)}(y)\) is the \(x^{th}\) derivative of Riemann Zeta function at \(y\).

Digvijay Singh - 1 year, 2 months ago

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