Question

How do you determine whether the sequence `a_n=n/(ln(n)^2` converges, if so how do you find the limit?

Answer 1

The series:

`sum_(n=2)^oo n/(lnn)^2`

is divergent.

Explanation:

Note: The question is not mentioning it expressly, but in the answer, I assume the series starts from `n=2` as `a_n` is undefined for `n=1`

A necessary condition for any series to converge is Cauchy's condition that:

`lim_(n->oo) a_n=0`

To verify whether this condition is met, first we note that

`a_n = f(n)`

where `f(x) = x/ln(x)^2` is a continuous function for `x in (1,+oo)`

So if the limit `lim_(x->oo) f(x)` exixst it must be the same as `lim_(n->oo) a_n`

Now:

`lim_(x->oo) f(x) = lim_(x->oo) x/(lnx)^2`

is the inderminate form `oo/oo` and we can solve it using the l'Hospital's rule (twice):

`lim_(x->oo) x/(lnx)^2 = lim_(x->oo) 1/((2lnx)/x) = lim_(x->oo) x/(2lnx) = lim_(x->oo) 1/(2/x) = lim_(x->oo) x/2 = +oo`

Thus also:

`lim_(n->oo) n/(lnn)^2 = +oo`

which implies that the series is not convergent.

As it has positive terms, the series is then divergent.