How to calculate the following sum:

$$\lim_{n\to\infty}\left(\frac{\ln n}3-\sum_{k=1}^n\frac1{3k-1}\right)=\lim_{n\to\infty}\left(\frac{\ln n}3-\sum_{k=1}^n\frac1{3k+2}\right)=\frac13\cdot\lim_{n\to\infty}\left(\ln n-\sum_{k=0}^{n-1}\frac1{k+\frac23}\right)$$

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.