# Convergence of $\sum_{n=2}^\infty \frac{1}{n^\alpha \ln^\beta (n)}$

Study the convergence of the following series: $$\sum_{n=2}^\infty \frac{1}{n^\alpha \cdot\ln^\beta(n)} \text{ where }\alpha,\beta \geq 0$$

Applying d'Alembert criterion I have that $$\lim_{n\to\infty} \frac{n^\alpha \ln^\beta(n)}{(n+1)^\alpha \ln^\beta (n+1)} = \lim_{n\to\infty} \left(\frac{n}{n+1}\right)^\alpha\left(\frac{\ln(n)}{\ln (n+1)}\right)^\beta = 1$$ so the nature of the series is inconclusive.

If $\alpha = \beta = 0$, then the series diverges, since $\sum_{n=2}^\infty 1 = \infty$.

Should I study the rest of the cases (i.e. if $\alpha = 0, \beta > 0$ the root test and the ratio test are also inconclusive). What is the best form to study the series?.

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You can apply Cauchy's condensation test. Moving to $$\sum_{n = 2}^{\infty} \frac{2^n}{2^{n\alpha} n^{\beta}} = \sum_{n = 2}^{\infty} \left( 2^{1-\alpha} \right)^n \frac{1}{n^{\beta}},$$ the series surrenders to the ratio test when $\alpha \neq 1$. When $\alpha=1$, this is a well-known series.
Sorry for commenting in an old post but I think that by the ratio test if $\alpha > 1$ the series converges. – user50554 Jan 28 '13 at 18:37
Yeah, which is consistent with the ratio test as $\frac{\left(2^{1-\alpha}\right)^{n+1}\frac{1}{\left(n+1\right)^{\beta}}}{\left(‌​2^{1-\alpha}\right)^{n}\frac{1}{n^{\beta}}} = 2^{1 - \alpha}\left(\frac{n}{n+1}\right)^{\beta} \to 2^{1 - \alpha} < 1$ if $\alpha > 1$. – levap Jan 28 '13 at 22:19
I think you should use the integral test. In order to establish the convergence/divergence of $$\int_2^{\infty} \frac{dx}{x^a\log^b{x}}$$ I suggest you to use the substitution $x=e^t$ which transforms the integral into $$\int_{\log 2}^{\infty} \frac{dt}{e^{(a-1)t}t^b}$$ which is easier. I think you should be able to conclude by yourself, now.