[Math] On convergence of Bertrand series $\sum\limits_{n=2}^{\infty} \frac{1}{n^{\alpha}\ln^{\beta}(n)}$ where $\alpha, \beta \in \mathbb{R}$

calculusconvergence-divergencelimitsreal-analysissequences-and-series

Study the convergence of $$\sum_{n=2}^{\infty}
\frac{1}{n^{\alpha}\ln^{\beta}(n)}$$ where $\alpha, \beta \in
\mathbb{R}$

I have proved that:

  1. This series diverges when $\alpha \leq 0$.

  2. This series converges when $\alpha > 1, \beta > 0$

  3. This series diverges when $0 < \alpha < 1, \beta > 0$

  4. This series converges when $\alpha = 1, \beta > 1$

Question: What happens when $\alpha > 0$ and $ \beta < 0$?

There are other questions on MSE which ask about this series, but this question is distinct because

  • I would like an argument which does not rely on the integral test for series convergence, and

  • this question considers all real $\alpha$ and $\beta$, while other questions ask only about $\alpha, \beta > 0$, where we can apply Cauchy condensation criterion

Best Answer

In that last case it diverges those are Bertrand's series see here: indeed we have

$$\frac{1}{n^{\alpha/2}\ln^\beta n} =\left[\frac{1}{\frac{2\beta}{\alpha}n^{\alpha/2\beta}\ln n^{\alpha/2\beta}}\right]^{\beta}\to \left[\frac{1}{\frac{2\beta}{\alpha}0^-}\right]^{\beta} =\infty $$ Since if $\alpha<0$ and $\beta>0$ then, $$ \lim_{n\to\infty}n^{\alpha/2\beta}=0\implies \lim_{n\to\infty}n^{\alpha/2\beta}\ln n^{\alpha/2\beta} =0^-$$

Then there exists $N$ such that $n>N$ we have

$$\frac{1}{n^{\alpha/2}\ln^\beta n}>1\implies \frac{1}{n^{\alpha}\ln^\beta n}>n^{-\alpha/2}$$ That is $$\sum_{n=N}^{\infty}\frac{1}{n^{\alpha}\ln^\beta n}>\sum_{n=N}^{\infty}n^{-\alpha/2} =\infty$$ from this you get the divergence

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