Let $p,q \in \mathbb{R}$

Consider :

$$\sum_{n=2}^\infty \frac{1}{n^p(\ln(n))^q}$$

For what $p,q$ does the series converge ?

Here's what I've got.

It suffices to check : $$\sum_{n=2}^\infty \frac{2^n}{(2^n)^p(\ln(2^n))^q} =\sum_{n=2}^\infty \frac{(2^{1-p})^n}{n^q} $$

So we require $ p>1$ or else the limit isnt 0 and it clearly diverges.

Now assuming $p>1 \implies |2^{1-p}| = z<1$

So by the Ratio Test $$\sum_{n=2}^\infty \frac{z^n}{n^q} $$

Converges $\forall q$

Hence the series converges if $p >1$

Is this right?


The series converges if $p > 1$, or if $p = 1$ and $q > 1$. It is possible to tackle both statements using condensation criteria, like you used (perhaps under a different name) in your proof.

It is also possible (perhaps even advisable) to use the integral test. So you are left trying to understand when $$ \int_2^\infty \frac{1}{x^p \ln^q x} dx $$ converges. The $p > 1$ case can be handled by direct comparison with $\dfrac{1}{x^p}$. The $p = 1$ case is where the integral test really shines.

  • $\begingroup$ Thanks Alot. If possible could I possibly trouble you for one last question. What would you say to ?$$\sum_{n=3}^{\infty}\frac{1}{n^p(ln(n))^q(ln(ln(n)))^r}$$ $\endgroup$ – Exc Mar 19 '15 at 4:16
  • $\begingroup$ I would again use the integral test. This is $u$ substitution, just like the first question. If you added another $\ln \ln \ln n$ on the bottom, it would be yet another, and so on. $\endgroup$ – davidlowryduda Mar 19 '15 at 4:22

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