2

Find the limit of $$\lim _{n\to \infty }\left(\sum _{k=1}^n\:\frac{\left[C^k_n\cdot a\right]}{C^k_{2n}}\right)$$ where $a$ is a real number, and [] denotes the integer part.

Solution:

I used the integer part inequility: $x-1< [x]\le x$.

Then we have that:

$$\left(\sum _{k=1}^n\:\frac{aC^k_n-1}{C^k_{2n}}\right)<\left(\sum _{k=1}^n\:\frac{\left[C^k_n\cdot a\right]}{C^k_{2n}}\right)\le\left(\sum _{k=1}^n\:\frac{C^k_n}{C^k_{2n}}\right)a$$ If I manage to prove that $$\lim _{n\to \infty }\left(\sum _{k=1}^n\:\frac{C^k_n}{C^k_{2n}}\right)a=\lim _{n\to \infty }\left(\sum _{k=1}^n\:\frac{aC^k_n-1}{C^k_{2n}}\right)=l$$

Then by sandwich theorem we will have that $$\lim _{n\to \infty }\left(\sum _{k=1}^n\:\frac{\left[C^k_n\cdot a\right]}{C^k_{2n}}\right)=l$$ But now I'm stuck.

shangq_tou
  • 998
  • You can do some cancelling (assuming "$C$" denotes binomial coefficients); for example, $k!$ would appear in both numerator and denominator – FShrike Aug 11 '21 at 09:27
  • That is what I get after cancelling: $\frac{C^k_n}{C^k_{2n}}=\frac{n!\left(2n-k\right)!}{\left(2n\right)!\left(n-k\right)!}$ – shangq_tou Aug 11 '21 at 09:30

1 Answers1

1

First, we split $a$ into its integral and fractional part $$ \left\lfloor {\left( \matrix{ n \cr k \cr} \right)a} \right\rfloor = \left\lfloor {\left( \matrix{ n \cr k \cr} \right) \left( {\left\lfloor a \right\rfloor + \left\{ a \right\}} \right)} \right\rfloor = \left( \matrix{ n \cr k \cr} \right)\left\lfloor a \right\rfloor + \left\lfloor {\left( \matrix{ n \cr k \cr} \right) \left\{ a \right\}} \right\rfloor \quad \left| {\;0 \le \left\{ a \right\} < 1} \right. $$ so that we get $$ \left( \matrix{ n \cr k \cr} \right)\left\lfloor a \right\rfloor \le \left\lfloor {\left( \matrix{ n \cr k \cr} \right)a} \right\rfloor < \left( \matrix{ n \cr k \cr} \right)\left( {\left\lfloor a \right\rfloor + 1} \right) $$

Then we rewrite the ratio of the binomials as $$ \eqalign{ & {1 \over {\left( \matrix{ 2n \cr k \cr} \right)}}\left( \matrix{ n \cr k \cr} \right) = {{n^{\,\underline {\,k\,} } } \over {k!{{\left( {2n} \right)^{\,\underline {\,k\,} } } \over {k!}}}} = {{n^{\,\underline {\,k\,} } } \over {\left( {2n} \right)^{\,\underline {\,k\,} } }} = \cr & = {{n^{\,\underline {\,k\,} } } \over {\left( {2n} \right)^{\,\underline {\,n + k - n\,} } }} = {{n^{\,\underline {\,k\,} } } \over {\left( {2n} \right)^{\,\underline {\,n\,} } n^{\,\underline {\,k - n\,} } }} = \cr & = {{n^{\,\underline {\,k\,} } } \over {\left( {2n} \right)^{\,\underline {\,n\,} } n^{\,\underline {\,k\,} } \left( {n - k} \right)^{\,\underline {\, - n\,} } }} = {1 \over {\left( {2n} \right)^{\,\underline {\,n\,} } \left( {n - k} \right)^{\,\underline {\, - n\,} } }} = \cr & = {{\left( {n - k + 1} \right)^{\,\overline {\,n\,} } } \over {\left( {2n} \right)^{\,\underline {\,n\,} } }} = {{\left( {2n - k} \right)^{\,\underline {\,n\,} } } \over {\left( {2n} \right)^{\,\underline {\,n\,} } }} = {1 \over {\left( \matrix{ 2n \cr n \cr} \right)}}\left( \matrix{2n - k \cr n \cr} \right) \cr} $$ where $x^{\,\underline {\,k\,} } ,\quad x^{\,\overline {\,k\,} } $ represent respectively the Falling and Rising Factorial

So the sum becomes $$ \eqalign{ & S(n) = \sum\limits_{k = 1}^n {{{\left( \matrix{ n \cr k \cr} \right)} \over {\left( \matrix{ 2n \cr k \cr} \right)}}} = \sum\limits_{k = 0}^n {{{\left( \matrix{ n \cr k \cr} \right)} \over {\left( \matrix{ 2n \cr k \cr} \right)}}} - 1 = \cr & = \left( {{1 \over {\left( \matrix{ 2n \cr n \cr} \right)}} \sum\limits_{0 \le \,k\;\left( { \le \,n} \right)} {\left( \matrix{ 2n - k \cr n \cr} \right)} } \right) - 1 = \cr & = {1 \over {\left( \matrix{ 2n \cr n \cr} \right)}}\left( \matrix{2n + 1 \cr n + 1 \cr} \right) - 1 = {1 \over {\left( \matrix{ 2n \cr n \cr} \right)}}{{2n + 1} \over {n + 1}} \left( \matrix{2n \cr n \cr} \right) - 1 = \cr & = {n \over {n + 1}} \cr} $$ and you can easily conclude about the limit

G Cab
  • 35,272