Question

How do you apply the ratio test to determine if `Sigma 1/(lnn)^n` from `n=[2,oo)` is convergent to divergent?

Answer 1

he series:

`sum_(n=2)^oo 1/(lnn)^n`

is convergent.

Explanation:

We have the series:

`sum_(n=2)^oo 1/(lnn)^n`

Now evaluate the ratio:

`abs(a_(n+1)/a_n) = abs ( (1/(ln(n+1)^(n+1)) ) /(1/(lnn)^n)) = (lnn)^n/(ln(n+1)^(n+1)) = (lnn/ln(n+1))^n 1/ln(n+1)`

Now consider the function:

`f(x) = lnx/ln(x+1)`

the limit for `x->oo` is in the form `oo/oo` so we can calculate it using l'Hospital's rule:

`lim_(x->oo) lnx/ln(x+1) = lim_(x->oo) (d/dx lnx)/(d/dx ln(x+1)) = lim_(x->oo) (1/x)/(1/(x+1)) = lim_(x->oo) (x+1)/x =1`

As `f(n) = lnn/ln(n+1)` we then have:

`lim_(n->oo) lnn/ln(n+1) =1`

and therefore:

`lim_(n->oo) (lnn/ln(n+1))^n =1`

Then:

`lim_(n->oo) abs(a_(n+1)/a_n) = lim_(n->oo) (lnn/ln(n+1))^n 1/ln(n+1) = lim_(n->oo) 1/ln(n+1) = 0`

which proves the series to be convergent.