How to determine whether the infinite series converges?

#sum_(x=0)^oo 1/(x*ln(x))^2#

1 Answer
Mar 27, 2018

Converges by the Comparison Test.

Explanation:

First, I am going to assume the series does not start at zero, as zero is not in the domain of the sequence. Moreover, #ln(0)# DNE. Also, the series should be in terms of #n,# integers, not #x,# I believe.

We'll let

#a_n=1/(nln(n))^2#, #b_n=1/n^2>a_n#, as removing the squared logarithm from the denominator ultimately results in a smaller denominator and therefore a larger sequence overall.

Now, we know

#sum_(n=1)^oo1/n^2# converges. It's the p-series #sum_(n=1)^oo1/n^p# with #p=2>1.#

So, since the larger series #sum_(n=1)^oob_n# converges, so must the smaller series #sum_(n=1)^ooa_n=sum_(n=1)^oo1/(nln(n))^2#, as per the Comparison Test.