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

Mar 9, 2018

${\lim}_{n \to \infty} {a}_{n} = 0$

#### Explanation:

Given:

${a}_{n} = {\left(\ln n\right)}^{5} / {n}^{\frac{1}{5}}$

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 $\frac{1}{5}$th power in the denominator might look like it would make a difference, but ${n}^{\frac{1}{25}} \to \infty$ faster than $\ln n \to \infty$.

• Hence ${\lim}_{n \to \infty} {a}_{n} = 0$

More rigorous analysis

Note that:

${a}_{{10}^{5 k}} = {\left(\ln {10}^{5 k}\right)}^{5} / \left({\left({10}^{5 k}\right)}^{\frac{1}{5}}\right) = {\left(5 k \ln 10\right)}^{5} / \left({\left({10}^{5 k}\right)}^{\frac{1}{5}}\right) = {\left(5 \ln 10\right)}^{5} {k}^{5} / {10}^{k}$

When $k \ge 3$ we find:

${a}_{{10}^{5 \left(k + 1\right)}} / {a}_{{10}^{5 k}} = \frac{{\left(5 \ln 10\right)}^{5} {\left(k + 1\right)}^{5} / {10}^{k + 1}}{{\left(5 \ln 10\right)}^{5} {k}^{5} / {10}^{k}}$

$\textcolor{w h i t e}{{a}_{{10}^{5 \left(k + 1\right)}} / {a}_{{10}^{5 k}}} = \frac{1}{10} {\left(\frac{k + 1}{k}\right)}^{5}$

$\textcolor{w h i t e}{{a}_{{10}^{5 \left(k + 1\right)}} / {a}_{{10}^{5 k}}} \le \frac{1}{10} {\left(\frac{4}{3}\right)}^{5} = \frac{1024}{2430} < \frac{1}{2}$

So by the ratio test:

${\lim}_{k \to \infty} {a}_{{10}^{5 k}} = 0$

Hence by monotonicity in the limit:

${\lim}_{n \to \infty} {a}_{n} = 0$