How do you apply the ratio test to determine if #sum_(n=2)^oo 10^n/(lnn)^n# is convergent to divergent?

Question

How do you apply the ratio test to determine if #sum_(n=2)^oo 10^n/(lnn)^n#  is convergent to divergent?

Paisley Johnson · Accepted Answer

The Ratio Test can be used to show that this series converges.

Let #a_{n}=10^{n}/((ln(n))^{n})#. Then #a_{n+1}/a_{n}=10^{n+1}/((ln(n+1))^{n+1}) * ((ln(n))^{n})/10^{n}# #=10/ln(n+1) * (ln(n)/ln(n+1))^{n}# Note that #(ln(n)/ln(n+1))^{n} <= 1# for all #n geq 2# (L'Hopital's Rule can be used to help show that the limit of this expression is 1, but it is enough to note that it is bounded). All this implies that #a_{n+1}/a_{n}=10/ln(n+1) * (ln(n)/ln(n+1))^{n}->0=L < 1# as #n->infty#. Therefore, the Ratio Test implies that #sum_{n=2}^{infty}a_{n}=sum_{n=2}^{infty}10^{n}/((ln(n))^{n})# converges.

Axel Alvarez · Answer

undefined To apply the ratio test to determine if the series ∑(n=2)^∞ 10^n / (lnn)^n is convergent or divergent, we first compute the limit of the absolute value of the ratio of successive terms as n approaches infinity.

Let's denote the nth term of the series as a_n = 10^n / (lnn)^n. Then, we calculate the ratio of consecutive terms:

r_n = |a_(n+1) / a_n| = |(10^(n+1) / (ln(n+1))^(n+1)) / (10^n / (lnn)^n)|

Simplify the expression:

r_n = |10^(n+1) / (ln(n+1))^(n+1) * (lnn)^n / 10^n|

= |10 / (ln(n+1) / lnn)^(n+1)|

= |10 * (lnn / ln(n+1))^(n+1)|

Now, we take the limit as n approaches infinity:

lim (n→∞) |10 * (lnn / ln(n+1))^(n+1)|

Since (lnn / ln(n+1))^(n+1) approaches 1 as n approaches infinity, the limit simplifies to:

lim (n→∞) |10| = 10

Since the limit of the absolute value of the ratio is greater than 1, by the ratio test, the series ∑(n=2)^∞ 10^n / (lnn)^n is divergent.