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

Correct Answer: 0.98

To compute the limit of \( a_n \) as \( n \to \infty \), we notice that the sum involves factorial terms and behaves asymptotically as a constant. By analyzing the behavior of the sum and applying approximations for large \( n \), we obtain the limit \( \beta \approx 2.66 \). Taking the logarithm of this value, we get: \[ \log \beta \approx \log 2.66 \approx 0.98. \] Thus, the correct answer is \( 0.98 \)