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For what values of $p$ does the following integral converge:

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

Ans. (Integral Test) $\int\limits_{n=2}^{n=\infty}\frac{1}{n(\ln n)^p} = \frac{1}{(-p+1)(ln\ n)^{p-1}}$

I know that $p \neq 1$, but I do not understand why the answer is $p > 1$

Joseph
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2 Answers2

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If $p < 1$, then the exponent of $\ln n$ is negative, which means you really have contribution from $\ln n$ in the numerator, to some positive power (namely to the $1 - p$ power). As $\ln n \to \infty$, for any positive $\alpha > 0$, we have $(\ln n)^\alpha \to \infty$. And that's why you have divergence when $p \leq 1$.

davidlowryduda
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  • Why is $\frac{1}{(ln\ n)^a}$ converges when $a > 0$? – Joseph Jan 27 '16 at 07:24
  • Because $\ln n \to \infty$, and dividing $1$ into very very many pieces means those pieces are small (i.e. going to zero). – davidlowryduda Jan 27 '16 at 07:27
  • Yes, but doesn't $(ln\ n)^a$ affect the convergence in anyway? For example, when $0 < a < 1$ – Joseph Jan 27 '16 at 07:29
  • I'm not sure what you're asking. Do you mean to have $(\ln n)^a$ not in a denominator right now? Regardless, $\ln n \to \infty$. So if you have it to a positive power $a$, then $(\ln n)^a \to \infty$. If you have a negative power $b$, then $(\ln n)^b \to 0$. The first is because positive powers of large numbers are large. The second is because dividing by positive powers of large numbers is like dividing by a large number, and is small. – davidlowryduda Jan 27 '16 at 07:37
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You can use the Cauchy condensation test. Therefore $$\sum_{n=2}^{\infty} \frac{1}{n(\ln\ n)^p}=\sum_{n=2}^{\infty}2^n \frac{1}{2^n(\ln\ 2^n)^p}=\sum_{n=2}^{\infty} \frac{1}{n^p(\ln2)^p}$$ that converges for $p>1$

Domenico Vuono
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