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How to prove that:

$$\lim_{n\rightarrow\infty}n\cdot\left(\log\left(3\right)-1-2\sum_{k=1}^{3^n}\frac{1}{\left(3k\right)^3-3k}\right)=0$$

when we know

$$\lim_{n\rightarrow\infty}\left(1+2\sum_{k=1}^n\frac{1}{\left(3k\right)^3-3k}\right)=\lim_{n\rightarrow\infty}\left(\sum_{k=1}^{3n+1}\frac{1}{k}-\sum_{k=1}^n\frac{1}{k}\right)=\log\left(3\right)$$ the demonstration of this equality is given by:

$$S=\lim_{n\rightarrow\infty}\left(\sum_{k=1}^{3n+1}\frac{1}{k}-\sum_{k=1}^n\frac{1}{k}\right)=\lim_{n\rightarrow\infty}\left(\sum_{k=1}^{3n+1}\frac{1}{k}-\sum_{k=1}^n\frac{1}{k}+\log\left(3n\right)-\log\left(3n\right)\right)$$ Reordering the terms $$S=\lim_{n\rightarrow\infty}\left(\frac{1}{3n+1}+\left(\sum_{k=1}^{3n}\frac{1}{k}-\log\left(3n\right)\right)-\left(\sum_{k=1}^n\frac{1}{k}-\log\left(n\right)\right)+\log\left(3\right)\right)$$ and we know that

$$\gamma=\lim_{n\rightarrow\infty}\left(\sum_{k=1}^n\frac{1}{k}-\log\left(n\right)\right)$$

and applying this up

$$ S=\lim_{n\rightarrow\infty}\left(\frac{1}{3n+1}+\gamma-\gamma+\log\left(3\right)\right)$$ So $$S=\log\left(3\right)$$

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    $\begingroup$ Answer is immediate via Cesaro-Stolz. Try it. $\endgroup$ – Paramanand Singh Oct 31 '17 at 18:27
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There missing link is the residual.

$S_m=\sum_\limits{k=1}^{3m+1}\frac{1}{k}-\sum_\limits{k=1}^{m}\frac{1}{k}= \sum_\limits{k=m+1}^{3m+1}\frac{1}{k}=\frac{1}{3m}\sum_\limits{k=m+1}^{3m+1}\frac{1}{(\frac{k}{3m})}$

The right hand side is a Riemann sum, so we get

$S_m = \int_\frac{1}{3}^1 \frac{1}{x} \mathrm{d}x + O(\frac{1}{m})= log(3)+O(\frac{1}{m})$

So

$LHS = \lim_{n\rightarrow\infty}n\cdot O(\frac{1}{3^n})=0$

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  • $\begingroup$ what does $O()$ mean? $\endgroup$ – Juan Alfaro Nov 2 '17 at 3:01
  • $\begingroup$ the big O notation $\endgroup$ – lion Nov 2 '17 at 3:04

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