# Test for convergence (2)

#### Alexmahone

##### Active member
Test for convergence: $\sum_2^\infty\frac{1}{n(\ln n)^p}$ when $p<0$.

My working:

Consider the function $f(x)=\frac{1}{x(\ln x)^p}=\frac{(\ln x)^q}{x}$ when $q=-p>0$.

In order to use the integral test, I have to establish that $f(x)$ is decreasing for $x\ge 2$. How do I proceed?

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#### Also sprach Zarathustra

##### Member
Test for convergence: $\sum_2^\infty\frac{1}{n(\ln n)^p}$ when $p<0$.

My working:

Consider the function $f(x)=\frac{1}{x(\ln x)^p}=\frac{(\ln x)^q}{x}$ when $q=-p>0$.

In order to use the integral test, I have to establish that $f(x)$ is decreasing for $x\ge 2$. How do I proceed?

Investigate when $\int_2^{\infty} f(x)dx$ converges (for which values $p$?).

If you have troubles with integration by parts, in this case, you can use:http://www.encyclopediaofmath.org/index.php/Ermakov_convergence_criterion

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#### Alexmahone

##### Active member
Investigate when $\int_2^{\infty} f(x)dx$ converges (for which values $p$?).
Actually, it's easier to use the comparison test:

If $p<0$, $\frac{1}{n(\ln n)^p}=\frac{(\ln n)^{-p}}{n}>\frac{1}{n}$ for $n\ge 3$.

Since $\sum_2\frac{1}{n}$ diverges, $\sum_2\frac{1}{n(\ln n)^p}$ also diverges.