Hi
Evaluation of \( \displaystyle \lim_{n\rightarrow \infty}\bigg[\binom{n}{0}\cdot \binom{n}{1}\cdot \binom{n}{2}\cdots \cdots \cdots \binom{n}{n}\bigg]^{\frac{1}{n(n+1)}} \)
Let \( L=\displaystyle \lim_{n\rightarrow \infty}\bigg[\binom{n}{0}\cdot \binom{n}{1}\cdot \binom{n}{2}\cdots \cdots \cdots \binom{n}{n}\bigg]^{\frac{1}{n(n+1)}} \).
Then:
\( log(L)=\displaystyle \lim_{n\rightarrow \infty} \dfrac{a_n}{b_n} \)
with:
\( a_n=\displaystyle\sum_{k=0}^nlog\binom{n}{k},\qquad b_n=n(n+1) \)
You can apply the
Soltz-Cesàro Theorem. Note that:
\( \displaystyle\dfrac{\displaystyle\binom{n+1}{k}}{\displaystyle\binom{n}{k}}=\dfrac{\dfrac{(n+1)!}{k!(n+1-k)!}}{\dfrac{n!}{k!(n-k)!}}=\dfrac{n+1}{n+1-k} \)
and from this:
\( c_n=a_{n+1}-a_n=\displaystyle\sum_{k=0}^nlog\left(\dfrac{n+1}{n+1-k}\right)=(n+1)log(n+1)-\displaystyle\sum_{k=1}^{n+1}log(k) \)
\( d_n=b_{n+1}-b_n=2(n+1) \)
and:
\( log(L)=\displaystyle \lim_{n\rightarrow \infty} \dfrac{a_n}{b_n}=\displaystyle \lim_{n\rightarrow \infty} \dfrac{c_n}{d_n} \)
You can appy the
Soltz-Cesàro Theorem again, where:
\( c_{n+1}-c_n=(n+2)log(n+2)-(n+1)log(n+1)-log(n+2)=(n+1)(log(n+2)-log(n+1))=log\left(\left(1+\dfrac{1}{n+1}\right)^{n+1}\right) \)
\( d_{n+1}-d_n=2 \)
Finally:
\( log(L)=\displaystyle \lim_{n\rightarrow \infty} \dfrac{log\left(\left(1+\dfrac{1}{n+1}\right)^{n+1}\right)}{2}=\dfrac{log(e)}{2}=1/2 \)
and \( L=e^{1/2} \).
Best regards.
