# y_n=log x_n, n =2,3,4,...and y_n-(n-1)/n y_(n-1)=1/n log n, with y_2=log sqrt2, how do you prove that x_n=(n!)^(1/n)?

Jan 24, 2017

By induction

#### Explanation:

Note that as ${y}_{n} = \log \left({x}_{n}\right)$, we have ${x}_{n} = {e}^{{y}_{n}}$ (assuming the natural logarithm. Using a different base will not change the structure of the proof, however). Additionally, note that ${y}_{n} = \frac{1}{n} \log \left(n\right) + \frac{n - 1}{n} {y}_{n - 1}$ by the second given relation.

Proof: (By induction)

Base Case: For $n = 2$, we are given ${y}_{2} = \log \left(\sqrt{2}\right)$, meaning

x_2 = e^log(sqrt(2)) = sqrt(2) = (2!)^(1/2)

Inductive Hypothesis: Suppose that x_k = (k!)^(1/k) for some integer $k \ge 2$.

Induction Step: We wish to show that x_(k+1) = [(k+1)!]^(1/(k+1)). Indeed, examining ${y}_{k + 1}$, we have

${y}_{k + 1} = \frac{1}{k + 1} \log \left(k + 1\right) + \frac{k}{k + 1} {y}_{k}$

$= \frac{1}{k + 1} \left[\log \left(k + 1\right) + k \log \left({x}_{k}\right)\right]$

=1/(k+1)[log(k+1)+klog((k!)^(1/k))]

=1/(k+1)[log(k+1)+log(k!)]

=1/(k+1)log(k!(k+1))

=1/(k+1)log((k+1)!)

=log([(k+1)!]^(1/(k+1)))

meaning x_(k+1) = e^(y_(k+1)) = [(k+1)!]^(1/(k+1)), as desired.

We have supposed true for $k$ and shown true for $k + 1$, thus, by induction, x_n = (n!)^(1/n) for all integers $n \ge 2$.