Using the integral test, how do you show whether #sum1 / (n (log n)^p ) # diverges or converges from n=3 to infinity?
1 Answer
For
Explanation:
I am assuming that
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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
If
#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
# = lim_(t->oo) int_3^t 1 / u "d"u#
# = lim_(t->oo) [ ln abs(u) ]_(x=3)^(x=t)#
... substitute back
# = 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
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2) Now let
Now let us take a look at the integral if
#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
# = 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
# = 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
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
# lim_(t-> oo) 1 / ((1-p) (ln t)^(p-1)) = 0# .
Thus, the integral is finite which means that for