Evaluate $\lim _{n \rightarrow \infty} \sum_{i=1}^{n} \frac{1}{i-3 n}$ Evaluate $\lim _{n \rightarrow \infty} \sum_{i=1}^{n} \frac{1}{i-3 n}$
Here $\sum_{i=1}^{n} \frac{1}{i-3 n}=\frac{1}{1-3 n}+\frac{1}{2-3 n}+\cdots+\frac{1}{n-3 n}$.
I tried to make the sum squeezed
between Convergent Sequences, but failed by getting
$\frac{-1}{2} \leq \frac{1}{-3 n+1}+\frac{1}{-3 n+2}+\cdots+\frac{1}{-3 n+n} \leq \frac{-1}{3}$

One approach is to use an expansion of harmonic numbers where $H(n)= \sum \limits_{i=1}^n \frac1n = \log_e(n) + \gamma + \frac{1}{2n}+O(n^{-2})$
You then have $\sum\limits_{i=1}^{n} \frac{1}{i-3 n} = H(2n-1)-H(3n-1) = \log_e(\frac{2n-1}{3n-1}) + \frac{n}{2(2n-1)(3n-1)}+O(n^{-2})$
and so $\lim\limits _{n \rightarrow \infty} \sum\limits_{i=1}^{n} \frac{1}{i-3 n} =\log_e(\frac{2}{3}) =\log_e(2)-\log_e(3)  \approx-0.405$
