Question

For n ∈ $\N$, let
$a_n=\frac{1}{n^{n-1}}\sum\limits_{k=0}^n\frac{n!}{k!(n-k)!}\frac{n^k}{k+1}$
and $\beta=\lim\limits_{n\rightarrow \infin}a_n$. Then, the value of log 𝛽 equals ___________ (rounded off to two decimal places).

Answer 1

The correct answer is 0.98 to 1.02. (approx)