Limit of a harmonic subseries minus "its" logarithm

Question

$\displaystyle \lim_{n \to \infty} \sum_{k=1}^n \frac{1}{3k-1} - \frac{1}{3}\ln(n)$

I think that inserting the other terms and then subtracting them would not help. I need just the ideea. Thank you.

Answer 1

It is not very difficult to show that, for natural values of a, $\displaystyle{\sum_{k=1}^n}\frac1{n+a}=H_{n+a}-H_a$ , where

$H_n$ is the $n^{th}$ harmonic number. But in our case, $a=-\frac13$ , so we have to extend the formula

or definition of $H_n$ so as to encompass non-natural arguments as well. Luckily for us, Euler

has already done it some $250$ years ago: $\displaystyle{H_n=\int_0^1\frac{1-x^n}{1-x}dx}$ . Using this, we can now compute

the value of $H_{-\frac13}$ to be $\displaystyle{\frac12\bigg(\frac\pi{\sqrt3}-3\ln3\bigg)}$. Now use the fact that $\displaystyle{\lim_{n\to\infty}H_n-\ln n}=\gamma$. See here

for more details on how to handle the integral in question.

Answer 2

If it were $\sum_{k=1}^n \frac 1k-\ln n$, we'd end up with $\gamma$. So I'd suggest to investigate $$ \sum_{k=1}^\infty\left(\frac 3{3k-1}-\frac1k\right)$$