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: $\gamma = \sum_{n=1}^{\infty} \left(\dfrac{1}{n} - \ln\left(1 + \dfrac{1}{n} \right) \right)$

#Calculus

Easy Math Editor

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Comments

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 - 4 years, 8 months ago

how can l find these solution

Liu Yang - 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Sum of ln⁡(n)n2\frac{\ln(n)}{n^2}n2ln(n)​

Comments

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