How are these related?

Algebra Level 5

$\displaystyle \lim _{ n\rightarrow \infty }{ \sqrt [ n ]{\binom{2n}{n-1} \binom{2n}{n} \binom{2n}{n+1} } }$

Determine the limit above.

As a challenge, do it without calculus.

The answer is 64.

1 solution

Dylan Pentland
May 19, 2015

Use that $\binom{2n}{n}$ is the largest value $\binom{2n}{k}$ can achieve, and that $\sum _{ k=0 }^{ 2n }{ \binom{2n}{n} } ={ 2 }^{ { 2 }n }$ . You know from the first fact that $\binom{2n}{n}\ge\frac { { 2 }^{ 2n } }{ 2n+1 }$ , because it's the largest term and cannot be below the average. We now have $\displaystyle \frac { { 2 }^{ 2n } }{ 2n+1 } \le \binom{2n}{n} \le {2}^{2n}$ Taking the nth roots, the expression becomes $\displaystyle 4 \left( \frac { 1 }{ \sqrt [ n ]{ 2n+1 } } \right) \le \sqrt [ n ]{ \binom{2n}{n} } \le 4$ As $n\rightarrow \infty$ , $\frac { 1 }{ \sqrt [ n ]{ 2n+1 } }$ becomes one. We now have that as $n$ goes to infinity $\displaystyle 4 \le \sqrt [ n ]{ \binom{2n}{n} } \le 4$ Therefore the limit is $4$ . Changing it to $\binom{2n}{n\pm1}$ will have no effect on the limit, since we can still make the assertion that it cannot be below average (since there are infinite terms). The expression comes out to ${4}^{3}$ or $64$ .

Moderator note:

Wonderful. Since the challenge is fulfilled, can you find the calculus approach?

The calculus approach to this problem would be to use an approximation for $\binom{2n}{n}$ as n goes to infinity. Probably the easiest method is to use Stirling's approximation for $n!$ , which says that $\displaystyle n!\sim \sqrt { 2\pi n } { \left( \frac { n }{ e } \right) }^{ n }$ as n goes to infinity. Then, use that $\binom{2n}{n}=\frac {(2n!)}{n!n!}$ . From this, we just use the formula and simplify to get $\displaystyle \binom{2n}{n} \sim \frac { { 4 }^{ n } }{ \sqrt { \pi n } }$ as n goes to infinity. When taking the nth root, $\sqrt [ n ]{ \sqrt { \pi n } }$ becomes 1 as $n$ goes to infinity and the top becomes 4.

Dylan Pentland - 6 years ago

oh god , I typed answer as 4 , forget to cube it !

Anyway , I used calculus , but your approach is nice ++1 .

I did in this way..... ${ P }_{ n }={ \left( \begin{matrix} 2n \\ n \end{matrix} \right) }^{ \cfrac { 1 }{ n } }\\ \lim _{ n\rightarrow \infty }{ \ln { { P }_{ n } } } =\lim _{ n\rightarrow \infty }{ \cfrac { 1 }{ n } \times \ln { \left( \cfrac { 2n.2n-1.2n-2........n+1 }{ n.n-1.n-2..............1 } \right) } } \\ L=\lim _{ n\rightarrow \infty }{ \cfrac { 1 }{ n } \sum _{ r=1 }^{ n }{ \ln { \left( \cfrac { 2n-r }{ n-r } \right) } } }$

now using Riemann sums

Nishu sharma - 6 years ago

How are these related?

The answer is 64.

1 solution

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