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