Show that: $$\sum_{k=2}^{n}\dfrac{\ln{k}}{k^2}\approx \ln{n}\cdot\left(\zeta_{n}{(2)}-\dfrac{\pi^2}{6}\right)+C,n\to\infty$$ where $$\zeta_{n}{(k)}=\sum_{j=1}^{n}\dfrac{1}{j^k}$$and $C$ is real constant.

I know this $$\zeta{(x)}=\sum_{n=1}^{\infty}\dfrac{1}{n^x}$$ $$\Longrightarrow \zeta'{(x)}=-\sum_{n=1}^{\infty}\dfrac{\ln{n}}{n^x}$$ let $x=2$,then we have $$\sum_{n=1}^{\infty}\dfrac{\ln{n}}{n^2}=\sum_{n=2}^{\infty}\dfrac{\ln{n}}{n^2}=-\zeta'{(2)}$$ But for this approximation,I can't prove it.

Thank you very much

  • $\begingroup$ $\zeta{(2)} = \pi^2/6$. It looks like you have to take the sum of the series minus the n first terms to achieve the result $\endgroup$ – T_O Apr 8 '14 at 8:10
  • $\begingroup$ Yes.@T_O,I know this,Thank you $\endgroup$ – china math Apr 8 '14 at 8:12
  • $\begingroup$ Don't mind my obvious suggestion, I didn't read it with enough attention :) $\endgroup$ – T_O Apr 8 '14 at 8:18
  • $\begingroup$ if it's the approximation you want, did you try Euler-Maclaurin formula? $\endgroup$ – Alex Apr 8 '14 at 8:42
  • $\begingroup$ I don't have an answer. Did you try any method related to mean value theorem? The resulting formula looks like it $\endgroup$ – wonghang Apr 8 '14 at 8:57

Rewriting the sums a bit, you're asking how to show that

$$ \sum_{k=2}^{\infty} \frac{\ln k}{k^2} - \sum_{k=n+1}^{\infty} \frac{\ln k}{k^2} \approx -\ln n \sum_{k=n+1}^{\infty} \frac{1}{k^2} + C. $$

Looking at this one might guess that we'll probably have $C = \sum_{k=2}^{\infty} \frac{\ln k}{k^2}$, so we would just need to prove that

$$ \sum_{k=n+1}^{\infty} \frac{\ln k}{k^2} \approx \ln n \sum_{k=n+1}^{\infty} \frac{1}{k^2}. \tag{1} $$

The argument you would use depends on what exactly you mean by $\approx$. For example, to leading order

$$ \sum_{k=n+1}^{\infty} \frac{\ln k}{k^2} \approx \int_n^\infty \frac{\ln x}{x^2}\,dx = \frac{\ln n + 1}{n} \approx \frac{\ln n}{n} $$


$$ \sum_{k=n+1}^{\infty} \frac{1}{k^2} \approx \int_n^\infty \frac{dx}{x^2} = \frac{1}{n}, $$

so we have a proof of $(1)$ in this sense.

  • $\begingroup$ I'm not sure what the downvoter has an issue with, but if you're seeking a more rigorous statement of the result then this method will yield that $$\sum_{k=2}^{n}\dfrac{\ln{k}}{k^2} = C + \ln{n}\cdot\left(\zeta_{n}{(2)}-\dfrac{\pi^2}{6}\right) + \epsilon_n,$$ where $C = \sum_{k=2}^{\infty} \frac{\ln k}{k^2}$ and $$\epsilon_n = o\left(\ln{n}\cdot\left(\zeta_{n}{(2)}-\dfrac{\pi^2}{6}\right)\right)$$ as $n \to \infty$. $\endgroup$ – Antonio Vargas Apr 8 '14 at 15:49
  • $\begingroup$ it's not me,But maybe you solution some wrong,Thank you for you solution $\endgroup$ – china math Apr 8 '14 at 15:50
  • $\begingroup$ I don't see an error in it, at least :) $\endgroup$ – Antonio Vargas Apr 8 '14 at 15:51

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