# Using the integral test, how do you show whether sum1 / (n (log n)^p )  diverges or converges from n=3 to infinity?

Mar 7, 2016

For $p = 1$, the row diverges. For $p > 1$, the row converges.

#### Explanation:

I am assuming that $p$ is a natural number, $p \ge 1$, and that $\log n$ is the natural logarithm $\ln n$.

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The integral test states:

The row

${\sum}_{n = 3}^{\infty} \frac{1}{n {\left(\ln n\right)}^{p}}$

converges to a real number if and only if the integral

${\int}_{3}^{\infty} \frac{1}{x {\left(\ln x\right)}^{p}} \text{d} x$

is finite.

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1) Let's first consider $p = 1$.

If $p = 1$, we need to determine if the following integral is finite:

${\int}_{3}^{\infty} \frac{1}{x \ln x} \text{d"x = lim_(t->oo) int_3^t 1 / (x ln x) "d} x$

$= {\lim}_{t \to \infty} {\int}_{3}^{t} \frac{1}{\ln} x \cdot \frac{1}{x} \text{d} x$

... substitute $u = \ln x \text{ " => ("d"u)/("d"x) = 1 / x => "d"u = 1 / x * "d} x$

$= {\lim}_{t \to \infty} {\int}_{3}^{t} \frac{1}{u} \text{d} u$

$= {\lim}_{t \to \infty} {\left[\ln \left\mid u \right\mid\right]}_{x = 3}^{x = t}$

... substitute back $u = \ln x$...

$= {\lim}_{t \to \infty} {\left[\ln \left\mid \ln x \right\mid\right]}_{x = 3}^{x = t}$

$= {\lim}_{t \to \infty} \left(\ln \left\mid \ln t \right\mid - \ln \left\mid \ln 3 \right\mid\right)$

The second term is clearly finite. However, we know that

${\lim}_{t \to \infty} \ln t = \infty$,

thus it also follows that

${\lim}_{t \to \infty} \ln \left(\left\mid \ln t \right\mid\right) = \infty$.

We can conclude that for $p = 1$, the integral is infinite and thus, the row diverges.

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2) Now let $p > 1$.

Now let us take a look at the integral if $p > 1$:

${\int}_{3}^{\infty} \frac{1}{x {\left(\ln x\right)}^{p}} \text{d" x = lim_(t -> oo) int_3^t 1 / (x (ln x)^p) "d} x$

$= {\lim}_{t \to \infty} {\int}_{3}^{t} \frac{1}{\ln x} ^ p \cdot \frac{1}{x} \text{d} x$

... substitute $u = \ln x \text{ " => ("d"u)/("d"x) = 1 / x => "d"u = 1 / x * "d} x$

$= {\lim}_{t \to \infty} {\int}_{3}^{t} \frac{1}{u} ^ p \text{ d} u$

$= {\lim}_{t \to \infty} {\int}_{3}^{t} {u}^{- p} \text{ d} u$

$= {\lim}_{t \to \infty} {\left[\text{ "1/(-p+1) u^(-p+1) " }\right]}_{x = 3}^{x = t}$

... substitute back $u = \ln x$...

$= {\lim}_{t \to \infty} {\left[\text{ "1/(1-p) (ln x)^(1-p) " }\right]}_{x = 3}^{x = t}$

$= {\lim}_{t \to \infty} {\left[\frac{1}{\left(1 - p\right) {\left(\ln x\right)}^{p - 1}}\right]}_{x = 3}^{x = t}$

$= {\lim}_{t \to \infty} \left[\frac{1}{\left(1 - p\right) {\left(\ln t\right)}^{p - 1}} - \frac{1}{\left(1 - p\right) {\left(\ln 3\right)}^{p - 1}}\right]$

For $p > 1$, the second term is clearly finite.

Thus, the only thing left to do is determine the limit behaviour

${\lim}_{t \to \infty} \frac{1}{\left(1 - p\right) {\left(\ln t\right)}^{p - 1}}$.

As we know that $\ln t$ diverges for $t \to \infty$, we can conclude that

${\lim}_{t \to \infty} \frac{1}{\left(1 - p\right) {\left(\ln t\right)}^{p - 1}} = 0$.

Thus, the integral is finite which means that for $p > 1$, the row converges.