Question

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

Answer 1

`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`