Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?.

Thanks in advance

share|improve this question

2 Answers 2

up vote 3 down vote accepted

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.

share|improve this answer
    
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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.