How do you test the series #Sigma (n^n)/(lnn)^n# from #n=[2,oo)# by the ratio test?

Question

Scarlett Allison · Accepted Answer

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

It's really easier in this case to use the root test:

#a_n = n^n/(lnn)^n = (n/lnn)^n#

So that:

#lim_(n->oo) root(n)(a_n) = lim_(n->oo) root(n)( (n/lnn)^n) = lim_(n->oo) n/lnn = oo#

Thus the series is not convergent, and as it has positive terms, it is divergent:

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

Samantha Adkison · Answer

undefined To test the series Σ(n^n)/(ln(n))^n from n=2 to infinity using the ratio test:

1. Calculate the ratio of consecutive terms: $ \frac{a_{n+1}}{a_n} = \frac{(n+1)^{n+1}/(ln(n+1))^{n+1}}{n^n/(ln(n))^n} $.

2. Simplify the ratio: $ \frac{a_{n+1}}{a_n} = \frac{(n+1)^{n+1}}{n^n} \times \frac{(ln(n))^n}{(ln(n+1))^{n+1}} $.

3. Take the limit as n approaches infinity of the absolute value of this ratio.

4. If the limit is less than 1, the series converges. If the limit is greater than 1 or it doesn't exist, the series diverges.

So, the series Σ(n^n)/(ln(n))^n converges if $ \lim_{{n \to \infty}} \left| \frac{(n+1)^{n+1}}{n^n} \times \frac{(ln(n))^n}{(ln(n+1))^{n+1}} \right| < 1 $.