How do you find the limit of the sequence: #a_n=(lnn)^5/n^(1/5#?

1 Answer
Mar 9, 2018

Answer:

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

Explanation:

Given:

#a_n = (ln n)^5/n^(1/5)#

Quick and dirty analysis

  • The logarithm #ln n# grows slower than any positive power of #n# since the exponential function grows faster.

  • Raising #ln n# to the #5#th power in the numerator and #n# to a small #1/5#th power in the denominator might look like it would make a difference, but #n^(1/25) -> oo# faster than #ln n -> oo#.

  • Hence #lim_(n->oo) a_n = 0#

More rigorous analysis

Note that:

#a_(10^(5k)) = (ln 10^(5k))^5/((10^(5k))^(1/5)) = (5k ln 10)^5/((10^(5k))^(1/5)) = (5 ln 10)^5 k^5/10^k#

When #k >= 3# we find:

#a_(10^(5(k+1)))/a_(10^(5k)) = ((5 ln 10)^5 (k+1)^5/10^(k+1))/((5 ln 10)^5 k^5/10^k)#

#color(white)(a_(10^(5(k+1)))/a_(10^(5k))) = 1/10((k+1)/k)^5#

#color(white)(a_(10^(5(k+1)))/a_(10^(5k))) <= 1/10(4/3)^5 = 1024/2430 < 1/2#

So by the ratio test:

#lim_(k->oo) a_(10^(5k)) = 0#

Hence by monotonicity in the limit:

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