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

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.