# Thread: Limit with binomial Coefficients

1. ## Limit with binomial Coefficients

$Finding value of$

$\displaystyle \lim_{n\rightarrow \infty} n^{-2}\cdot \sum^{n}_{k=0}\ln\bigg(\binom{n}{k}\bigg)$

2. ## Re: Limit with binomial Coefficients

consider stirlings formula

3. ## Re: Limit with binomial Coefficients

Originally Posted by juantheron
$Finding value of$

$\displaystyle \lim_{n\rightarrow \infty} n^{-2}\cdot \sum^{n}_{k=0}\ln\bigg(\binom{n}{k}\bigg)$
The first and last index terms can be discarded, since log(1) = 0

Rearrange the factorials to obtain:

$\lim_{n \to \infty} \frac{1}{n^2} \sum_{k=1}^{n-1} \log{\binom{n}{k}} = \lim_{n \to \infty} \frac{1}{n^2} \sum_{k=1}^{n} \log{ \left( k^{2k-n-1} \right) }$

From there, an integral approximation of the sum should be sufficient and necessary to extract the limit.

4. ## Re: Limit with binomial Coefficients

Originally Posted by BenHowe
consider stirlings formula
slight overkill you can just use the thing i just posted with a riemann approximation to find the limit

5. ## Re: Limit with binomial Coefficients

To paradoxica would you like to explain me in detail. Thanks

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