7
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find the

$$\lim_{n\to+\infty}\left(\dfrac{\ln{2^2}}{2^2}+\dfrac{\ln{3^2}}{3^2}+\dfrac{\ln{4^2}}{4^2}+\cdots+\dfrac{\ln{n^2}}{n^2}\right)$$

My try: $$\lim_{n\to+\infty}\left(\dfrac{\ln{2^2}}{2^2}+\dfrac{\ln{3^2}}{3^2}+\dfrac{\ln{4^2}}{4^2}+\cdots+\dfrac{\ln{n^2}}{n^2}\right)=2\sum_{n=2}^{\infty}\dfrac{\ln{n}}{n^2}$$

and I know solve this following

$$\sum_{n=2}^{\infty}(-1)^n\dfrac{\ln{n}}{n}=\ln{2}\left(C-\dfrac{\ln{2}}{2}\right)$$

where $C$ is Euler constant

Solution:note this following $$\lim_{n\to\infty}\left(\dfrac{\ln{1}}{1}+\dfrac{\ln{2}}{2}+\cdots+\dfrac{\ln{n}}{n}-\dfrac{(\ln{n})^2}{2}\right)=l$$ we let $$S_{n}=\sum_{k=1}^{n}\dfrac{(-1)^k\ln{k}}{k}$$

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  • $\begingroup$ Are you sure it's $\ln (i^2)$ and not $(\ln i)^2$, right? $\endgroup$ – leonbloy Oct 22 '13 at 11:35
  • $\begingroup$ It's $\ln{(i^2)}$ $\endgroup$ – math110 Oct 22 '13 at 11:40
  • $\begingroup$ There is some discussion at oeis.org/A073002, including an expression in terms of the Glaisher-Kinkelin constant (whatever that is). $\endgroup$ – Gerry Myerson Oct 22 '13 at 11:51
  • $\begingroup$ Thank you,@GerryMyerson,But My answer How have this $\endgroup$ – math110 Oct 22 '13 at 11:54
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With the Riemann zeta function for $x>1$ $$\zeta(x) = \sum_{n=1}^{\infty}\frac{1}{n^x}$$ you get, taking the derivative term-wise $$\zeta'(x) = -\sum_{n=1}^{\infty}\frac{\ln{n}}{n^x},$$ so your sum is (the first term vanishes) $$2\sum_{n=2}^{\infty}\dfrac{\ln{n}}{n^2} = -2\zeta'(2) = 1.8750965\dots$$

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  • $\begingroup$ Thank you,maybe have other close form? $\endgroup$ – math110 Oct 22 '13 at 11:42
  • $\begingroup$ Thank you,My book is answer it's $$\dfrac{1}{3}\pi^2(\gamma+\ln{(2\pi)}-12\ln{A})$$,where $A$ is Glaisher-Kinkelin constant $\endgroup$ – math110 Oct 22 '13 at 11:53
  • $\begingroup$ Yes, see mathworld.wolfram.com/RiemannZetaFunction.html formula (45), but IMO you missed a - sign. $\endgroup$ – gammatester Oct 22 '13 at 11:55
  • $\begingroup$ How get this,Thank you $\endgroup$ – math110 Oct 22 '13 at 11:59

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