limit of sums of integer parts 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.
 A: 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
