# Limit with binomial Coefficients (1 Viewer)

#### juantheron

##### Active Member
$\bg_white Finding value of$

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

Last edited:

#### BenHowe

##### Active Member
consider stirlings formula

##### -insert title here-
$\bg_white Finding value of$

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

Rearrange the factorials to obtain:

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

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

##### -insert title here-
consider stirlings formula
slight overkill you can just use the thing i just posted with a riemann approximation to find the limit

#### juantheron

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

#### tywebb

##### dangerman
$\bg_white Finding value of$

$\bg_white \displaystyle \lim_{n\rightarrow \infty} n^{-2}\cdot \sum^{n}_{k=0}\ln\bigg(\binom{n}{k}\bigg)$
Code:
Use $$blah blah$$ instead of $blah blah$ to avoid Invalid Equation message.
$\bg_white \displaystyle \lim_{n\rightarrow \infty} n^{-2}\cdot \sum^{n}_{k=0}\ln\bigg(\binom{n}{k}\bigg)$

Put into wolframalpha and you get 0.5