# studying the series $\sum_\limits{n=1}^\infty \frac{1}{n^ {\alpha}(\log n)^ {\beta}}$. [duplicate]

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 $\log n<n$

## marked as duplicate by Mark Viola real-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Mar 16 '18 at 15:49

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 $\beta>1$
• @Anne You are welcome! Bye – user Mar 16 '18 at 16:55

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 $\displaystyle\sum\dfrac{1}{n^{\alpha-\xi}}<\infty$.

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).