Compute $\sum_{n\geq 1}u_n$ where $u_n=\frac{1}{n}$ if $n\not\equiv 0\pmod 3$ and $\frac{-2}{n}$ if $n\equiv 0\pmod 3$.

Question

Compute $\sum_{n\geq 1}u_n$ where $$u_n=\begin{cases} \frac{1}{n}&n\not\equiv 0\pmod 3\\ \frac{-2}{n}&n\equiv 0\pmod 3\end{cases}.$$

I really tried many thing, but not concludive. I set $$v_p=\frac{1}{3p-2}+\frac{1}{3p-2}-\frac{2}{3p}.$$ Therefore $$\sum_{k\geq 1}^{3n}u_k=\sum_{k=1}^n v_k=\sum_{k=1}^n\frac{1}{3k}+\sum_{k=1}^n\frac{1}{3k-1}+\sum_{k=1}^n\frac{1}{3k-2}.$$ Then I tried to get the same denominator, but I gut stuck because the $3k-1$ and $3k-2$.

Answer 1

Hint

You also have $$v_p=\frac{1}{3k-2}+\frac{1}{3k-1}+\frac{1}{3k}-\frac{3}{3k},$$ i.e. $$\sum_{k=1}^n v_k=\sum_{k=1}^{3n}\frac{1}{k}-\sum_{k=1}^n\frac{1}{k}.$$

Answer 2

You mean $$\sum_{k=1}^{3n}u_k=\sum_{k=1}^n\frac1{3k-2} +\sum_{k=1}^n\frac1{3k-1}-2\sum_{k=1}^n\frac1{3k} =H_{3n}-H_n$$ where $H_n=\sum_{k=1}^n1/k$. Then $H_n=\ln n+\gamma+o(1)$ and so $$\sum_{k=1}^{3n}u_k=\ln 3+o(1)$$ etc.