An Euler-Mascheroni-like sequence

Question

How does one compute the limit of the sequence:

$$\sum_{k = 0}^{n}\frac{1}{3k+1} - \frac{\ln(n)}{3}$$

I would apreciate a hint.

Answer 1

Hint: use a comparison series/integral, by writing $$\frac{\ln n}{3} = \int_{1}^n \frac{dx}{3x}=\sum_{k=1}^{n-1} \int_{k}^{k+1}\frac{dx}{3x}$$ and $$\sum_{k=0}^n \frac{1}{3k+1} = \sum_{k=0}^n \int_{k}^{k+1}\frac{dx}{3k+1}$$

Edit: is not a straightforward approach -- I was thinking about proving convergence.