Convergence of series with integral test

Question

Given that the following series is convergent, determine the values of p.

$$\sum_{n=2}^{\infty}\dfrac{1}{n(\log(n))^p}$$

So far what I have done is using the integral test, in order to use integral test, I set p $\in(0, \infty)$ to make f(x) decreasing.

Therefore I got:

$$\int_{2}^{\infty}f(x)dx=\dfrac{1}{(p-1)[\log(2)]^{1-p}}$$ which exist for $p\neq1$.

So my answer is $p\in(1,\infty)$, perhaps my question is stupid (sorry about that), but am I on the right track, didn't I make any mistakes?

thanks

You're right. A series $\sum\limits_{n\ge2}\dfrac1{ n^a\log^bn}$ is known as a Bertrand's series, and it is convergent if $a>1$ or $a=1, \enspace b>1$. — Bernard, May 01 '15 at 23:21
somehow I have no idea about why $\int log(x) dx = \int ln(x) dx$ @Bernard — Alana, May 04 '15 at 00:01
thats exactly my question as well, since the answer seems missing a constant $ln(10)$ for changing base ? — Alana, May 04 '15 at 00:10
I think they use the old name of the function (still is use for the logarithm of a complex number) – which, by the way, was Log. — Bernard, May 04 '15 at 00:16

Adhvaitha · Accepted Answer · 2015-05-02T00:19:11.540

1

The integral to be compared with is $$\int_2^{\infty}\dfrac{dx}{x\ln^p(x)} = \int_{\ln2}^{\infty} \dfrac{dt}{t^p} = \left. \dfrac{t^{1-p}}{1-p}\right \vert_{\ln2}^{\infty} = - \dfrac{(\ln2)^{1-p}}{1-p} + \lim_{t\to \infty} \dfrac{t^{1-p}}{1-p}$$ The limit exists only for $p>1$ and doesn't exist for $p \leq 1$. Hence, the series $$\sum_{n=2}^{\infty} \dfrac1{n\ln^p(n)}$$ converges for $p>1$ and diverges for $p \leq 1$.

edited May 02 '15 at 00:19

answered May 01 '15 at 23:23

Adhvaitha

20,259

How did you go from the first to second step? I can't see it. – Shinaolord May 02 '15 at 00:04
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@user57404 Make the substitution $t=\ln(x)$ and see what happens. – Adhvaitha May 02 '15 at 00:07
Ah okay, I was definitely looking too hard. Thought it'd be a complicated sub. Wow. – Shinaolord May 02 '15 at 00:15
somehow I have no idea about why $\int log(x) dx = \int ln(x) dx$ – Alana May 04 '15 at 00:00
@Alan For mathematician, $\log$ means $\ln$. For non-mathematician, typically $\log$ denotes $\log_{10}$. – Adhvaitha May 04 '15 at 00:06

Convergence of series with integral test

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