Question

`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)`?

Answer 1

By induction

Explanation:

Note that as `y_n = log(x_n)`, 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 = 1/nlog(n)+(n-1)/ny_(n-1)` by the second given relation.

Proof: (By induction)

Base Case: For `n=2`, we are given `y_2 = log(sqrt(2))`, 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>=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) = 1/(k+1)log(k+1)+k/(k+1)y_k`

`=1/(k+1)[log(k+1)+klog(x_k)]`

`=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>=2`.
∎