How to deal with convergence of $\sum_{n=1}^{\infty} (-1)^{ \lfloor{\frac{n^3+n+1}{3n^2-1}} \rfloor } \cdot \frac{\log(n)}{n} $

Question

I have a small problem to check convergence of expression like that: $$\sum_{n=1}^{\infty}u_n= \sum_{n=1}^{\infty} (-1)^{\left \lfloor{\frac{n^3+n+1}{3n^2-1}}\right \rfloor } \cdot \frac{\log(n)}{n} $$ On my lessons usually we have situations when $\sum_{n=1} \left| u_n \right|$ converges and then from the criterion of absolute convergence $ \sum_{n=1}^{\infty}u_n $ converges too. Ok, it is simple. Other situation is when I have something with regular behavior for example $ (-1)^n \cdot sth $ But there I have totally hard to predict and catch some rules expression and how can I check convergence of that? Of course the sum can be conditionally convergent but what next to be done there?

Answer 1

• A first simplification: Observing that $$\frac{n^3+n+1}{3n^2-1} = \frac{n}{3}+o(1)$$ we have $$\left\lfloor \frac{n^3+n+1}{3n^2-1} \right\rfloor = \left\lfloor\frac{n}{3}\right\rfloor$$ for $n$ large enough. Therefore, it is enough to study $$\sum_n (-1)^{\left\lfloor\frac{n}{3}\right\rfloor} \frac{\log n}{n}$$

• To deal with this (obviously, as you noted we do not have absolute convergence), note that the sign is constant for triples: $3k,3k+1,3k+2$. I.e., $$ \sum_{n=3}^{3N} (-1)^{\left\lfloor\frac{n}{3}\right\rfloor} \frac{\log n}{n} = \sum_{k=1}^{N} (-1)^{k} \underbrace{\sum_{n=3k}^{3k+2}\frac{\log n}{n}}_{v_k} $$ Can you use some theorem you know to study the convergence of $\sum_{k}(-1)^k v_k$?

  (For instance, first you may want to show that $(v_k)_k$ is decreasing.)