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

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Answer 1 · 2017-03-05T02:06:21

he series:

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

is convergent.

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.

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