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"Find the values of $p$ s.t. the following series converges: $\sum_{n=2}^{\infty} \frac{1}{n^p \ln(n)}$"

I am trying to do this problem through using the Integral Test to find the values of $p$. I know that for $p = 0$, the series diverges so I will only be considering values of $p \neq 0$.

The function $f(x) = (x^p \ln(x))^{-1}$ satisfies the criterion of the Integral Test, but I am having a difficult time integrating the function.

We have $$\int_{n=2}^{\infty} \frac{1}{x^p \ln(x)} dx$$ A u-substitution with $u = \ln(x)$ will not help us and neither would setting it to $x^p$. Someone suggested letting $x = e^u$, but...i'm not so sure where we would go with that.

Martin Sleziak
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Ozera
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4 Answers4

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  • For $p=1$ use test integral.
  • For $0<p<1$, choose $\ell \in (p,1)$, and remark that $n^{\ell}\dfrac{1}{n^p\ln n}\to +\infty$.
  • For $p>1$, $\dfrac{1}{n^p\ln n}\leq \dfrac{1}{n^p}$ for $n>>>$.
    For the case $0<p<1$ let $\ell\in (p,1)$, then $\dfrac{n^{\ell}}{n^p\ln n}=\dfrac{n^{\ell-p}}{\ln n}\to +\infty$ (because $\ell-p>0$), hence for $n>>>$, $\dfrac{n^{\ell}}{n^p\ln n}\geq 1$ i.e $\dfrac{1}{n^p\ln n}\geq \dfrac{1}{n^{\ell}}$, but $\sum\dfrac{1}{n^{\ell}}$ diverge, we get the result.
Hamou
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  • eh? There is no clever substitution I can use for the integral? – Ozera Aug 22 '14 at 03:53
  • For $p=1$, $\left(\ln(\ln x)\right)'=\dfrac{1}{x\ln x}$. – Hamou Aug 22 '14 at 03:55
  • Could explain in detail your hint for the case when $ 0 < p < 1$ ? I understand the other two cases now. Thanks. – Ozera Aug 22 '14 at 05:43
  • @Ozera This case is edited in the answer. – Hamou Aug 22 '14 at 06:25
  • Hmm, could you tell me if my explanation works? I didn't compare it to the harmonic series though.

    For the case 0 < p < 1, lets choose an l s.t. l \in (p, 1) for n^l. The motivation for this is to help us choose a series which is smaller than our original.

    We may then compare $1/(x^p ln(x))$ to $1/n^l$. Since we have $1/n^l < 1/(x^p ln(x))$ and $1/n^l$ is a divergent p-series, our original series is also divergent (to +infinity) by the Direct-Comparison test.

     – Ozera Aug 22 '14 at 07:15
  • Yes, that's it. Right – Hamou Aug 22 '14 at 07:29
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Another approach is Cauchy condensation theorem which says the original series converges if and only if the series

$$ \sum_{n}\frac{2^n}{n 2^{np}} $$

does. See here.

Community
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Mhenni Benghorbal
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    Ah...I suppose I could try this method as a last resort. Though, I am hoping someone will give advice that makes the problem easier to solve. – Ozera Aug 22 '14 at 04:04
  • @Ozera: As I said it is a diferent approach to the problem and at the same time it is a nice theorem to know. – Mhenni Benghorbal Aug 22 '14 at 04:08
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For simplicity I ignore the term $n=2$ in the beginning because $\log(2) < 1$.

Prove first that $\sum_{n=3}^\infty \frac{1}{n \log(n)}$ diverges using Cauchy condensation test or the integral test.

Then if $p > 1$, we have $\sum_{n=3}^\infty \frac{1}{n^p \log(n)} \le \sum_{n=3}^\infty \frac{1}{n^p} < \infty$.

If $0 \le p \le 1$, then $\sum_{n=3}^\infty \frac{1}{n^p \log(n)} \ge \sum_{n=3}^\infty \frac{1}{n \log(n)} = \infty$.

So the series converges if and only if $p > 1$.

J. J.
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Comparison test is best here. But here's how to use that substitution idea. Notice how it comes down to comparison in the end.

Using $x=e^u$ we have $dx=e^u du$. Thus $$\int_2^\infty \frac{dx}{x^p \ln x} = \int_{\ln2}^\infty \frac{e^u}{u e^{p u}} du = \int_{\ln2}^\infty e^{(1-p) u} \frac{1}{u} du.$$ It is easy to see that for $p<1$ the integrand is unbounded, so we have divergence. For $p=1$ the antiderivative is known and we again diverge. For $p>1$ we can bound the integrand above by $e^{(1-p)u}$, and the integral of this is easily found to be convergent.

zibadawa timmy
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