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).
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).
\(a_n=\frac{1}{n^{n-1}}\sum\limits_{k=0}^n\frac{n!}{k!(n-k)!}\frac{n^k}{k+1}\)
and \(\beta=\lim\limits_{n\rightarrow \infin}a_n\). Then, the value of log 𝛽 equals ___________ (rounded off to two decimal places).
Updated On: Oct 28, 2024
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Correct Answer: 0.98
Solution and Explanation
The correct answer is 0.98 to 1.02. (approx)
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