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I seem to be struggling with a mathematical problem. I need to evaluate this:

enter image description here

I know that the sum of all natural logarithms is ln(n!) and that the sum of all natural cubes is (n(n+1)/2)^2 but these don't seem to be of any help in this situation. I would really appreciate some help!

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  • $\begingroup$ In general: $\sum _{i=1}^n \log ^m(i)=\underset{x\to 0}{\text{lim}}\frac{\partial ^mH_n^{(-x)}}{\partial x^m}$ where:$H_n^{(-x)}$ is Harmonic Number of order x. $\endgroup$ Mar 14 '20 at 13:06
  • $\begingroup$ @MariuszIwaniuk and what is ∂ then? $\endgroup$
    – KapitaiN
    Mar 14 '20 at 13:23
  • $\begingroup$ Symbol of derivative,I yours case m-th derivative.Answer from CAS like Mathematica: $$\zeta ^{(3,0)}(0,n+1)-\frac{3 \gamma _2}{2}-3 \gamma _1 (\gamma +\log (2 \pi ))+\zeta (3)+\frac{1}{8} \left(-8 \gamma ^3+4 \log ^3(2 \pi )-12 \gamma ^2 \log (2 \pi )+\pi ^2 \log (2 \pi )\right)$$ $\endgroup$ Mar 14 '20 at 13:48
  • $\begingroup$ "sum of all natural logarithms" ? You mean $$\sum_{j=1}^n \ln(j)=\ln(n!)$$ Similar for the sum of the cubes. $\endgroup$
    – Peter
    Mar 14 '20 at 14:26
  • $\begingroup$ @Peter your writing is correct but in my case the logarithm itself is cubed and not the j so I can't add them like that. $\endgroup$
    – KapitaiN
    Mar 14 '20 at 14:38
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I do not know if there is a closed form.

If you want to get at least an order of magnitude, consider $$\int_{i=1}^n \log^3(i) \,di < \sum_{i=1}^n \log^3(i) <\int_{i=1}^{n+1} \log^3(i) \,di $$ with $$\int \log^3(i) \,di= -6 i+i \log ^3(i)-3 i \log ^2(i)+6 i \log (i)$$

For $n=100$, the left integral is $5573.28$, the sum is $5622.17$ and the right integral is $5671.26$.

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