How may one calculate $$\lim_{n\to\infty} \ \left(\left(\sum_{k=1}^{n} \frac{1}{3k-1}\right) - \frac{\ln n}{3}\right) \ ?$$
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Manipulations based on the definition of the Euler Mascheroni constant, which is similar, $H_n - log(n)$, might help. – Torsten Hĕrculĕ Cärlemän Dec 22 '13 at 16:40
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You may use computational methods to find a sufficient limit for large n, however, I am still thinking of an exact value – Torsten Hĕrculĕ Cärlemän Dec 22 '13 at 16:48
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Generally speaking, $$\lim_{n\to\infty}\left(\ln n-\sum_{k=0}^n\frac1{k-a}\right)=\pi\cdot\cot(a\pi)+H_a-\gamma,$$ where $H_a=\displaystyle\int_0^1\frac{1-x^a}{1-x\ }dx$ is the generalized harmonic number, and $\gamma$ is the Euler-Mascheroni constant. In our case, $a=\dfrac13$ – Lucian Dec 22 '13 at 16:51
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@Lucian Could you please give a link to the statement? – Torsten Hĕrculĕ Cärlemän Dec 22 '13 at 16:53
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2I don't understand how this question is off-topic (there are currently 3 close votes for that reason). – Jonathan Dec 22 '13 at 17:19
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I agree. There seems to be a cabal of close-for-no-reason people. – marty cohen Aug 31 '19 at 00:47
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$$\lim_{n\to\infty}\left(\frac{\ln n}3-\sum_{k=1}^n\frac1{3k-1}\right)=\lim_{n\to\infty}\left(\frac{\ln n}3-\sum_{k=1}^n\frac1{3k+2}\right)=\frac13\cdot\lim_{n\to\infty}\left(\ln n-\sum_{k=0}^{n-1}\frac1{k+\frac23}\right)$$
Now, suppose that instead of $\dfrac23$ we have a natural number m. Knowing that $\displaystyle\lim_{n\to\infty}\bigg[\ln n-$
$-\displaystyle\sum_{k=1}^n\frac1k\bigg]=\gamma,~$ it is trivial to show that $\displaystyle\lim_{n\to\infty}\left[\ln n\!-\!\sum_{k=1}^n\frac1{k+m}\right]=\gamma+H_m,~$ where $H_m=$
$=\displaystyle\sum_{k=1}^m\frac1k~$ is the $m^{th}$ harmonic number. The only trouble is that $\dfrac23\not\in\mathbb{N}$, so the above definition
does not apply here in this particular case. Therefore, we must seek to find another one, whose
meaning can easily be extended to non-natural arguments as well. Luckily for us, Euler already
found it three centuries ago! Here it is: $H_a=\displaystyle\int_0^1\frac{1-x^a}{1-x~~}dx.~$ In our case, $a=\dfrac23.~$ I have
evaluated a similar one here, for $a'=\dfrac13.~$ Wikipedia also lists their values.
Lucian
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