Limit with binomial Coefficients (1 Viewer)

BenHowe · Jun 6, 2018

consider stirlings formula

Paradoxica · Jun 6, 2018

juantheron said:
$Finding value of$

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

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.

Paradoxica · Jun 6, 2018

BenHowe said:
consider stirlings formula

slight overkill you can just use the thing i just posted with a riemann approximation to find the limit

juantheron · Jun 7, 2018

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

tywebb · Mar 25, 2023

juantheron said:
$Finding value of$

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

Code:

Use [tex]blah blah[/tex] instead of $blah blah$ to avoid Invalid Equation message.

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

Put into wolframalpha and you get 0.5

Limit with binomial Coefficients (1 Viewer)

juantheron

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BenHowe

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Paradoxica

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juantheron

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tywebb

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