How to prove upper bound for partial sum of binomial coefficients












2












$begingroup$


I just come across this inequality for the upper bound of partial sum of binomial coefficients, that is
begin{array}
$sum_{k=0}^{m}binom{n}{k}leq (frac{en}{m})^{m},
end{array}
I have trying to prove it but not having much success. I have used the stirling forluma $n!=sqrt{2pi}n^{n+1/2}e^{^{-n+r(n)}}$, where $r(n)in (frac{1}{12n+1}, frac{1}{12n})$, thus I get $frac{1}{k!}leq (frac{e}{k})^{k}$, and $binom{n}{m}leq (frac{en}{m})^{m}$ for all intergers $m in [1, n]$, which is just one item case, but that still has a big gap with $sum_{k=0}^{m}binom{n}{k}leq (frac{en}{m})^{m}$.
I just don't how to further expand the case the partial sum case? Any hints or insights would be helpful! Thanks!






combinatorics inequality binomial-coefficients







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    I just come across this inequality for the upper bound of partial sum of binomial coefficients, that is
    begin{array}
    $sum_{k=0}^{m}binom{n}{k}leq (frac{en}{m})^{m},
    end{array}
    I have trying to prove it but not having much success. I have used the stirling forluma $n!=sqrt{2pi}n^{n+1/2}e^{^{-n+r(n)}}$, where $r(n)in (frac{1}{12n+1}, frac{1}{12n})$, thus I get $frac{1}{k!}leq (frac{e}{k})^{k}$, and $binom{n}{m}leq (frac{en}{m})^{m}$ for all intergers $m in [1, n]$, which is just one item case, but that still has a big gap with $sum_{k=0}^{m}binom{n}{k}leq (frac{en}{m})^{m}$.
    I just don't how to further expand the case the partial sum case? Any hints or insights would be helpful! Thanks!






    combinatorics inequality binomial-coefficients







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      I just come across this inequality for the upper bound of partial sum of binomial coefficients, that is
      begin{array}
      $sum_{k=0}^{m}binom{n}{k}leq (frac{en}{m})^{m},
      end{array}
      I have trying to prove it but not having much success. I have used the stirling forluma $n!=sqrt{2pi}n^{n+1/2}e^{^{-n+r(n)}}$, where $r(n)in (frac{1}{12n+1}, frac{1}{12n})$, thus I get $frac{1}{k!}leq (frac{e}{k})^{k}$, and $binom{n}{m}leq (frac{en}{m})^{m}$ for all intergers $m in [1, n]$, which is just one item case, but that still has a big gap with $sum_{k=0}^{m}binom{n}{k}leq (frac{en}{m})^{m}$.
      I just don't how to further expand the case the partial sum case? Any hints or insights would be helpful! Thanks!






      combinatorics inequality binomial-coefficients







      share|cite|improve this question











      $endgroup$




      I just come across this inequality for the upper bound of partial sum of binomial coefficients, that is
      begin{array}
      $sum_{k=0}^{m}binom{n}{k}leq (frac{en}{m})^{m},
      end{array}
      I have trying to prove it but not having much success. I have used the stirling forluma $n!=sqrt{2pi}n^{n+1/2}e^{^{-n+r(n)}}$, where $r(n)in (frac{1}{12n+1}, frac{1}{12n})$, thus I get $frac{1}{k!}leq (frac{e}{k})^{k}$, and $binom{n}{m}leq (frac{en}{m})^{m}$ for all intergers $m in [1, n]$, which is just one item case, but that still has a big gap with $sum_{k=0}^{m}binom{n}{k}leq (frac{en}{m})^{m}$.
      I just don't how to further expand the case the partial sum case? Any hints or insights would be helpful! Thanks!






      combinatorics inequality binomial-coefficients




      combinatorics inequality binomial-coefficients






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      share|cite|improve this question













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      edited Dec 17 '18 at 17:57









      Alex Ravsky

      41.5k32283




      41.5k32283










      asked Dec 17 '18 at 14:55









      JizheJizhe

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          $begingroup$

          For each $1le mle n$ put $$f(m,n)=sum_{k=0}^{m}binom{n}{k}mbox{ and }g(m,n)=left (frac{en}{m}right)^{m}.$$ We prove that $f(m,n)<g(m,n) $ by induction with respect to $n$. For $m=1$ we have $$f(m,n)=1+n<en=g(m,n).$$ For $n=m$ we have $f(m,n)=2^n<e^n$. In particular, we have the base of the induction. Using that $${nchoose k}+{nchoose k+1}={n+1choose k+1}$$ for each $0le kle n-1$, we have that $$f(m+1, n+1)=f(m,n)+f(m+1,n)$$ for each $1le mle n-1$. By the induction hypothesis, in order to show that $$f(m+1,n+1)<g(m+1,n+1)$$ it suffices to show that $$g(m+1, n+1)ge g(m,n)+g(m+1,n).$$ Check this inequality



          $$g(m+1, n+1)ge g(m,n)+g(m+1,n)$$



          $$left(frac{e(n+1)}{m+1}right)^{m+1}ge left (frac{en}{m}right)^{m}+left (frac{en}{m+1}right)^{m+1}$$



          $$eleft(frac{(n+1)}{m+1}right)^{m+1}ge left (frac{n}{m}right)^{m}+eleft (frac{n}{m+1}right)^{m+1}$$



          Since
          $$left (frac{(n+1)}{m+1}right)^{m+1}=left (frac{n}{m+1}+frac{1}{m+1}right)^{m+1}ge
          left (frac{n}{m+1}right)^{m+1}+(m+1) left(frac{n}{m+1}right)^{m}frac{1}{m+1},$$



          it suffices to show that



          $$eleft (frac{n}{m+1}right)^{m}ge left (frac{n}{m}right)^{m}$$



          $$ege left (frac{m+1}{m}right)^{m}$$



          $$e^{frac 1m}ge frac{m+1}{m}$$



          Which is true, because



          $$e^{frac 1m}=1+{frac 1m}+{frac 1{2m^2}}+dots.$$






          share|cite|improve this answer









          $endgroup$













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            $begingroup$

            For each $1le mle n$ put $$f(m,n)=sum_{k=0}^{m}binom{n}{k}mbox{ and }g(m,n)=left (frac{en}{m}right)^{m}.$$ We prove that $f(m,n)<g(m,n) $ by induction with respect to $n$. For $m=1$ we have $$f(m,n)=1+n<en=g(m,n).$$ For $n=m$ we have $f(m,n)=2^n<e^n$. In particular, we have the base of the induction. Using that $${nchoose k}+{nchoose k+1}={n+1choose k+1}$$ for each $0le kle n-1$, we have that $$f(m+1, n+1)=f(m,n)+f(m+1,n)$$ for each $1le mle n-1$. By the induction hypothesis, in order to show that $$f(m+1,n+1)<g(m+1,n+1)$$ it suffices to show that $$g(m+1, n+1)ge g(m,n)+g(m+1,n).$$ Check this inequality



            $$g(m+1, n+1)ge g(m,n)+g(m+1,n)$$



            $$left(frac{e(n+1)}{m+1}right)^{m+1}ge left (frac{en}{m}right)^{m}+left (frac{en}{m+1}right)^{m+1}$$



            $$eleft(frac{(n+1)}{m+1}right)^{m+1}ge left (frac{n}{m}right)^{m}+eleft (frac{n}{m+1}right)^{m+1}$$



            Since
            $$left (frac{(n+1)}{m+1}right)^{m+1}=left (frac{n}{m+1}+frac{1}{m+1}right)^{m+1}ge
            left (frac{n}{m+1}right)^{m+1}+(m+1) left(frac{n}{m+1}right)^{m}frac{1}{m+1},$$



            it suffices to show that



            $$eleft (frac{n}{m+1}right)^{m}ge left (frac{n}{m}right)^{m}$$



            $$ege left (frac{m+1}{m}right)^{m}$$



            $$e^{frac 1m}ge frac{m+1}{m}$$



            Which is true, because



            $$e^{frac 1m}=1+{frac 1m}+{frac 1{2m^2}}+dots.$$






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              For each $1le mle n$ put $$f(m,n)=sum_{k=0}^{m}binom{n}{k}mbox{ and }g(m,n)=left (frac{en}{m}right)^{m}.$$ We prove that $f(m,n)<g(m,n) $ by induction with respect to $n$. For $m=1$ we have $$f(m,n)=1+n<en=g(m,n).$$ For $n=m$ we have $f(m,n)=2^n<e^n$. In particular, we have the base of the induction. Using that $${nchoose k}+{nchoose k+1}={n+1choose k+1}$$ for each $0le kle n-1$, we have that $$f(m+1, n+1)=f(m,n)+f(m+1,n)$$ for each $1le mle n-1$. By the induction hypothesis, in order to show that $$f(m+1,n+1)<g(m+1,n+1)$$ it suffices to show that $$g(m+1, n+1)ge g(m,n)+g(m+1,n).$$ Check this inequality



              $$g(m+1, n+1)ge g(m,n)+g(m+1,n)$$



              $$left(frac{e(n+1)}{m+1}right)^{m+1}ge left (frac{en}{m}right)^{m}+left (frac{en}{m+1}right)^{m+1}$$



              $$eleft(frac{(n+1)}{m+1}right)^{m+1}ge left (frac{n}{m}right)^{m}+eleft (frac{n}{m+1}right)^{m+1}$$



              Since
              $$left (frac{(n+1)}{m+1}right)^{m+1}=left (frac{n}{m+1}+frac{1}{m+1}right)^{m+1}ge
              left (frac{n}{m+1}right)^{m+1}+(m+1) left(frac{n}{m+1}right)^{m}frac{1}{m+1},$$



              it suffices to show that



              $$eleft (frac{n}{m+1}right)^{m}ge left (frac{n}{m}right)^{m}$$



              $$ege left (frac{m+1}{m}right)^{m}$$



              $$e^{frac 1m}ge frac{m+1}{m}$$



              Which is true, because



              $$e^{frac 1m}=1+{frac 1m}+{frac 1{2m^2}}+dots.$$






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                For each $1le mle n$ put $$f(m,n)=sum_{k=0}^{m}binom{n}{k}mbox{ and }g(m,n)=left (frac{en}{m}right)^{m}.$$ We prove that $f(m,n)<g(m,n) $ by induction with respect to $n$. For $m=1$ we have $$f(m,n)=1+n<en=g(m,n).$$ For $n=m$ we have $f(m,n)=2^n<e^n$. In particular, we have the base of the induction. Using that $${nchoose k}+{nchoose k+1}={n+1choose k+1}$$ for each $0le kle n-1$, we have that $$f(m+1, n+1)=f(m,n)+f(m+1,n)$$ for each $1le mle n-1$. By the induction hypothesis, in order to show that $$f(m+1,n+1)<g(m+1,n+1)$$ it suffices to show that $$g(m+1, n+1)ge g(m,n)+g(m+1,n).$$ Check this inequality



                $$g(m+1, n+1)ge g(m,n)+g(m+1,n)$$



                $$left(frac{e(n+1)}{m+1}right)^{m+1}ge left (frac{en}{m}right)^{m}+left (frac{en}{m+1}right)^{m+1}$$



                $$eleft(frac{(n+1)}{m+1}right)^{m+1}ge left (frac{n}{m}right)^{m}+eleft (frac{n}{m+1}right)^{m+1}$$



                Since
                $$left (frac{(n+1)}{m+1}right)^{m+1}=left (frac{n}{m+1}+frac{1}{m+1}right)^{m+1}ge
                left (frac{n}{m+1}right)^{m+1}+(m+1) left(frac{n}{m+1}right)^{m}frac{1}{m+1},$$



                it suffices to show that



                $$eleft (frac{n}{m+1}right)^{m}ge left (frac{n}{m}right)^{m}$$



                $$ege left (frac{m+1}{m}right)^{m}$$



                $$e^{frac 1m}ge frac{m+1}{m}$$



                Which is true, because



                $$e^{frac 1m}=1+{frac 1m}+{frac 1{2m^2}}+dots.$$






                share|cite|improve this answer









                $endgroup$



                For each $1le mle n$ put $$f(m,n)=sum_{k=0}^{m}binom{n}{k}mbox{ and }g(m,n)=left (frac{en}{m}right)^{m}.$$ We prove that $f(m,n)<g(m,n) $ by induction with respect to $n$. For $m=1$ we have $$f(m,n)=1+n<en=g(m,n).$$ For $n=m$ we have $f(m,n)=2^n<e^n$. In particular, we have the base of the induction. Using that $${nchoose k}+{nchoose k+1}={n+1choose k+1}$$ for each $0le kle n-1$, we have that $$f(m+1, n+1)=f(m,n)+f(m+1,n)$$ for each $1le mle n-1$. By the induction hypothesis, in order to show that $$f(m+1,n+1)<g(m+1,n+1)$$ it suffices to show that $$g(m+1, n+1)ge g(m,n)+g(m+1,n).$$ Check this inequality



                $$g(m+1, n+1)ge g(m,n)+g(m+1,n)$$



                $$left(frac{e(n+1)}{m+1}right)^{m+1}ge left (frac{en}{m}right)^{m}+left (frac{en}{m+1}right)^{m+1}$$



                $$eleft(frac{(n+1)}{m+1}right)^{m+1}ge left (frac{n}{m}right)^{m}+eleft (frac{n}{m+1}right)^{m+1}$$



                Since
                $$left (frac{(n+1)}{m+1}right)^{m+1}=left (frac{n}{m+1}+frac{1}{m+1}right)^{m+1}ge
                left (frac{n}{m+1}right)^{m+1}+(m+1) left(frac{n}{m+1}right)^{m}frac{1}{m+1},$$



                it suffices to show that



                $$eleft (frac{n}{m+1}right)^{m}ge left (frac{n}{m}right)^{m}$$



                $$ege left (frac{m+1}{m}right)^{m}$$



                $$e^{frac 1m}ge frac{m+1}{m}$$



                Which is true, because



                $$e^{frac 1m}=1+{frac 1m}+{frac 1{2m^2}}+dots.$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 17 '18 at 20:53









                Alex RavskyAlex Ravsky

                41.5k32283




                41.5k32283






























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