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I have to study the character of this series $$\sum_{n=1}^\infty \frac{1}{n^{\alpha}(\log n)^{\beta}}.$$

$\alpha$ and $\beta$ are two parameters.

I'm considering the case $\alpha >1 $ and all real values for $\beta $ .

If $\beta \ge 0$, $\exists n_0 \in N $ such that $\forall n\ge n_0, \dfrac{1}{n^ {\alpha}(\log n)^ {\beta}}<\dfrac{1}{n^ {\alpha}}$.

For the comparison test $\dfrac{1}{n^{\alpha} (\log n)^ {\beta}}$ converges.

I tried to analyze the case $-1<\beta<0,\beta=-1, \beta<-1$ but I'm not sure.

Anyway, for $\beta<-1$ the series is convergent for $\alpha > 1- \beta$, divergent for $\alpha < 1- \beta$.

For $-1<\beta<0$ the series is convergent for $\alpha > 2$, divergent for $\alpha <2$.

For $\beta=-1$ the series is convergent for $\alpha > 2$, divergent for $1<\alpha < 2$.

I've used the fact that $\log n<n$

Lorenzo B.
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Anne
  • 2,969

3 Answers3

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You can refer to Cauchy condensation test

$$ 0 \ \leq\ \sum_{n=1}^{\infty} f(n)\ \leq\ \sum_{n=0}^{\infty} 2^{n}f(2^{n})\ \leq\ 2\sum_{n=1}^{\infty} f(n)$$

and study the equivalent condensed series

$$\sum_{n=1}^\infty \dfrac{2^n}{2^{n\alpha}(\log 2^n)^{\beta}}=\sum_{n=1}^\infty \dfrac{1}{2^{n(\alpha-1)}n^{\beta}\log^{\beta}2}$$

from which is clear that the series converges if and only if

  • $\alpha>1$
  • $\alpha=1$ and $\beta>1$
user
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For $\beta<0$, so we are dealing with $\dfrac{(\log n)^{\gamma}}{n^{\alpha}}$ for $\gamma>0$.

Now use the fact that $(\log n)^{\gamma}\leq C_{\gamma,\xi}n^{\xi}$ where $\alpha-\xi>1$, $\xi>0$, where $C_{\gamma,\xi}>0$ is some constant, then $\dfrac{(\log n)^{\gamma}}{n^{\alpha}}\leq C_{\gamma,\xi}\dfrac{1}{n^{\alpha-\xi}}$ and $\displaystyle\sum\dfrac{1}{n^{\alpha-\xi}}<\infty$.

user284331
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The results for this series (which incidentally is known as a Bertrand's series) are much simpler: it converges if and only if

  • $\alpha >1$ (by comparison with a Riemann series)
  • or $\alpha=1$ and $\beta>1$ (by the integral test).
Bernard
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