Question

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

Answer 1

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

Explanation:

I am assuming that `p` is a natural number, `p >= 1`, and that `log n` is the natural logarithm `ln n`.

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The integral test states:

The row

`sum_(n=3)^(oo) 1 / (n (ln n)^p)`

converges to a real number if and only if the integral

`int_3^(oo) 1 / (x (ln x)^p) "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^(oo) 1 / (x ln x) "d"x = lim_(t->oo) int_3^t 1 / (x ln x) "d"x`

`= lim_(t->oo) int_3^t 1 / ln x * 1 / x "d"x`

... substitute `u = ln x " " => ("d"u)/("d"x) = 1 / x => "d"u = 1 / x * "d" x`

`= lim_(t->oo) int_3^t 1 / u "d"u`

`= lim_(t->oo) [ ln abs(u) ]_(x=3)^(x=t)`

... substitute back `u = ln x`...

`= lim_(t->oo) [ ln abs(ln x) ]_(x=3)^(x=t)`

`= lim_(t->oo) ( ln abs (ln t) - ln abs(ln 3) )`

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

`lim_(t->oo) ln t = oo`,

thus it also follows that

`lim_(t->oo) ln (abs(ln t)) = oo`.

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^(oo) 1 / (x (ln x)^p) "d" x = lim_(t -> oo) int_3^t 1 / (x (ln x)^p) "d" x`

`= lim_(t->oo) int_3^t 1 / (ln x)^p * 1 / x "d"x`

... substitute `u = ln x " " => ("d"u)/("d"x) = 1 / x => "d"u = 1 / x * "d" x`

`= lim_(t->oo) int_3^t 1 / u^p " d" u`

`= lim_(t->oo) int_3^t u^(-p) " d" u`

`= lim_(t->oo) [ " "1/(-p+1) u^(-p+1) " "]_(x=3)^(x=t)`

... substitute back `u = ln x`...

`= lim_(t->oo) [ " "1/(1-p) (ln x)^(1-p) " "]_(x=3)^(x=t)`

`= lim_(t->oo) [ 1 / ((1-p) (ln x)^(p-1))]_(x=3)^(x=t)`

`= lim_(t->oo) [ 1 / ((1-p) (ln t)^(p-1)) - 1 / ((1-p) (ln 3)^(p-1)) ]`

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

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

`lim_(t-> oo) 1 / ((1 - p) (ln t)^(p-1))`.

As we know that `ln t` diverges for `t -> oo`, we can conclude that

`lim_(t-> oo) 1 / ((1-p) (ln t)^(p-1)) = 0`.

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