Suppose $lim_{n to infty}a_n=a$, Prove $lim_{nto infty} frac{1}{2^n} sum_{k=0}^n binom{n}{k}a_k=a$.












2












$begingroup$



Suppose $lim_{n to infty}a_n=a$. Prove that $$lim_{nto infty} frac{1}{2^n} sum_{k=0}^n binom{n}{k}a_k=a,.$$




Here is a proof,



Proof: Consider $t_{n,k}=frac{1}{2^n}binom{n}{k}$,$0 leq t_{n,k}= frac{n(n-1)...(n-k+1)}{2^nk(k-1)...1}leqfrac{n^k}{2^n} $ and by squeeze theorem, $lim_{n to infty}t_{n,k}=0$. Also, $sum_{k=0}^n frac{1}{2^n}binom{n}{k}=1$, so then by Silverman–Toeplitz theorem, we have $lim_{nto infty} frac{1}{2^n} sum_{k=0}^n binom{n}{k}a_k=a$.



Is this proof correct? Is there other cool proofs?






real-analysis sequences-and-series limits proof-verification alternative-proof







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What is topliz?
    $endgroup$
    – Batominovski
    Dec 14 '18 at 3:41






  • 1




    $begingroup$
    Ultimately this is just a consequence of the fact that $$frac 1 {2^n} sum_{k = 0}^n binom{n}{k} = 1$$ while the individual summands tend to zero, so this acts as an averaging operator. You may be interested in the ergodic theorem, which considers a similar class of averages.
    $endgroup$
    – T. Bongers
    Dec 14 '18 at 3:46










  • $begingroup$
    Sorry! It is Silverman–Toeplitz theorem, I miss spelled it :en.wikipedia.org/wiki/Silverman%E2%80%93Toeplitz_theorem
    $endgroup$
    – nafhgood
    Dec 14 '18 at 3:46










  • $begingroup$
    I'm sure many probabilistic proofs could be given given that this question could be cast as a sequence of random variables. You might draw in that attention with the correct tag.
    $endgroup$
    – Robert Wolfe
    Dec 14 '18 at 5:58
















2












$begingroup$



Suppose $lim_{n to infty}a_n=a$. Prove that $$lim_{nto infty} frac{1}{2^n} sum_{k=0}^n binom{n}{k}a_k=a,.$$




Here is a proof,



Proof: Consider $t_{n,k}=frac{1}{2^n}binom{n}{k}$,$0 leq t_{n,k}= frac{n(n-1)...(n-k+1)}{2^nk(k-1)...1}leqfrac{n^k}{2^n} $ and by squeeze theorem, $lim_{n to infty}t_{n,k}=0$. Also, $sum_{k=0}^n frac{1}{2^n}binom{n}{k}=1$, so then by Silverman–Toeplitz theorem, we have $lim_{nto infty} frac{1}{2^n} sum_{k=0}^n binom{n}{k}a_k=a$.



Is this proof correct? Is there other cool proofs?






real-analysis sequences-and-series limits proof-verification alternative-proof







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What is topliz?
    $endgroup$
    – Batominovski
    Dec 14 '18 at 3:41






  • 1




    $begingroup$
    Ultimately this is just a consequence of the fact that $$frac 1 {2^n} sum_{k = 0}^n binom{n}{k} = 1$$ while the individual summands tend to zero, so this acts as an averaging operator. You may be interested in the ergodic theorem, which considers a similar class of averages.
    $endgroup$
    – T. Bongers
    Dec 14 '18 at 3:46










  • $begingroup$
    Sorry! It is Silverman–Toeplitz theorem, I miss spelled it :en.wikipedia.org/wiki/Silverman%E2%80%93Toeplitz_theorem
    $endgroup$
    – nafhgood
    Dec 14 '18 at 3:46










  • $begingroup$
    I'm sure many probabilistic proofs could be given given that this question could be cast as a sequence of random variables. You might draw in that attention with the correct tag.
    $endgroup$
    – Robert Wolfe
    Dec 14 '18 at 5:58














2












2








2


0



$begingroup$



Suppose $lim_{n to infty}a_n=a$. Prove that $$lim_{nto infty} frac{1}{2^n} sum_{k=0}^n binom{n}{k}a_k=a,.$$




Here is a proof,



Proof: Consider $t_{n,k}=frac{1}{2^n}binom{n}{k}$,$0 leq t_{n,k}= frac{n(n-1)...(n-k+1)}{2^nk(k-1)...1}leqfrac{n^k}{2^n} $ and by squeeze theorem, $lim_{n to infty}t_{n,k}=0$. Also, $sum_{k=0}^n frac{1}{2^n}binom{n}{k}=1$, so then by Silverman–Toeplitz theorem, we have $lim_{nto infty} frac{1}{2^n} sum_{k=0}^n binom{n}{k}a_k=a$.



Is this proof correct? Is there other cool proofs?






real-analysis sequences-and-series limits proof-verification alternative-proof







share|cite|improve this question











$endgroup$





Suppose $lim_{n to infty}a_n=a$. Prove that $$lim_{nto infty} frac{1}{2^n} sum_{k=0}^n binom{n}{k}a_k=a,.$$




Here is a proof,



Proof: Consider $t_{n,k}=frac{1}{2^n}binom{n}{k}$,$0 leq t_{n,k}= frac{n(n-1)...(n-k+1)}{2^nk(k-1)...1}leqfrac{n^k}{2^n} $ and by squeeze theorem, $lim_{n to infty}t_{n,k}=0$. Also, $sum_{k=0}^n frac{1}{2^n}binom{n}{k}=1$, so then by Silverman–Toeplitz theorem, we have $lim_{nto infty} frac{1}{2^n} sum_{k=0}^n binom{n}{k}a_k=a$.



Is this proof correct? Is there other cool proofs?






real-analysis sequences-and-series limits proof-verification alternative-proof




real-analysis sequences-and-series limits proof-verification alternative-proof






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 14 '18 at 3:45







nafhgood

















asked Dec 14 '18 at 3:33









nafhgoodnafhgood

1,805422




1,805422








  • 1




    $begingroup$
    What is topliz?
    $endgroup$
    – Batominovski
    Dec 14 '18 at 3:41






  • 1




    $begingroup$
    Ultimately this is just a consequence of the fact that $$frac 1 {2^n} sum_{k = 0}^n binom{n}{k} = 1$$ while the individual summands tend to zero, so this acts as an averaging operator. You may be interested in the ergodic theorem, which considers a similar class of averages.
    $endgroup$
    – T. Bongers
    Dec 14 '18 at 3:46










  • $begingroup$
    Sorry! It is Silverman–Toeplitz theorem, I miss spelled it :en.wikipedia.org/wiki/Silverman%E2%80%93Toeplitz_theorem
    $endgroup$
    – nafhgood
    Dec 14 '18 at 3:46










  • $begingroup$
    I'm sure many probabilistic proofs could be given given that this question could be cast as a sequence of random variables. You might draw in that attention with the correct tag.
    $endgroup$
    – Robert Wolfe
    Dec 14 '18 at 5:58














  • 1




    $begingroup$
    What is topliz?
    $endgroup$
    – Batominovski
    Dec 14 '18 at 3:41






  • 1




    $begingroup$
    Ultimately this is just a consequence of the fact that $$frac 1 {2^n} sum_{k = 0}^n binom{n}{k} = 1$$ while the individual summands tend to zero, so this acts as an averaging operator. You may be interested in the ergodic theorem, which considers a similar class of averages.
    $endgroup$
    – T. Bongers
    Dec 14 '18 at 3:46










  • $begingroup$
    Sorry! It is Silverman–Toeplitz theorem, I miss spelled it :en.wikipedia.org/wiki/Silverman%E2%80%93Toeplitz_theorem
    $endgroup$
    – nafhgood
    Dec 14 '18 at 3:46










  • $begingroup$
    I'm sure many probabilistic proofs could be given given that this question could be cast as a sequence of random variables. You might draw in that attention with the correct tag.
    $endgroup$
    – Robert Wolfe
    Dec 14 '18 at 5:58








1




1




$begingroup$
What is topliz?
$endgroup$
– Batominovski
Dec 14 '18 at 3:41




$begingroup$
What is topliz?
$endgroup$
– Batominovski
Dec 14 '18 at 3:41




1




1




$begingroup$
Ultimately this is just a consequence of the fact that $$frac 1 {2^n} sum_{k = 0}^n binom{n}{k} = 1$$ while the individual summands tend to zero, so this acts as an averaging operator. You may be interested in the ergodic theorem, which considers a similar class of averages.
$endgroup$
– T. Bongers
Dec 14 '18 at 3:46




$begingroup$
Ultimately this is just a consequence of the fact that $$frac 1 {2^n} sum_{k = 0}^n binom{n}{k} = 1$$ while the individual summands tend to zero, so this acts as an averaging operator. You may be interested in the ergodic theorem, which considers a similar class of averages.
$endgroup$
– T. Bongers
Dec 14 '18 at 3:46












$begingroup$
Sorry! It is Silverman–Toeplitz theorem, I miss spelled it :en.wikipedia.org/wiki/Silverman%E2%80%93Toeplitz_theorem
$endgroup$
– nafhgood
Dec 14 '18 at 3:46




$begingroup$
Sorry! It is Silverman–Toeplitz theorem, I miss spelled it :en.wikipedia.org/wiki/Silverman%E2%80%93Toeplitz_theorem
$endgroup$
– nafhgood
Dec 14 '18 at 3:46












$begingroup$
I'm sure many probabilistic proofs could be given given that this question could be cast as a sequence of random variables. You might draw in that attention with the correct tag.
$endgroup$
– Robert Wolfe
Dec 14 '18 at 5:58




$begingroup$
I'm sure many probabilistic proofs could be given given that this question could be cast as a sequence of random variables. You might draw in that attention with the correct tag.
$endgroup$
– Robert Wolfe
Dec 14 '18 at 5:58










2 Answers
2






active

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1












$begingroup$

Say $|a_n| le C$ for each $n$. For $epsilon > 0$, take $N$ s.t. $|a_n-a| < epsilon$ for $n ge N$. Then for $M ge N^2$, $$frac{1}{2^M}sum_{k=0}^M {M choose k}|a_k-a| le frac{1}{2^M}sum_{k=0}^sqrt{M} {M choose k}2C+frac{1}{2^M}sum_{k=sqrt{M}}^M {M choose k}epsilon.$$ So, $limsup_{M to infty} |frac{1}{2^M}sum_{k=0}^M {M choose k} a_k - a| le epsilon$. This holds for each $epsilon > 0$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    There's a pretty standard exercise in which we are asked to show that if $a_krightarrow a$ then
    $$biggl(frac{1}{n}sum_{k=1}^n a_kbiggr)rightarrow a,.$$
    It's reasonable to suspect that a similar approach there might work here.



    After one starts down this road, you eventually realize that you'll need to show that




    $$lim_{nrightarrowinfty}frac{1}{2^n}sum_{k=0}^Nbinom{n}{k}=0$$




    where $N$ is some possibly-big, but-nevertheless-fixed positive integer. To accomplish this, we notice
    $$sum_{k=0}^Nbinom{n}{k}leqsum_{k=0}^Nfrac{n^k}{k!}=sum_{k=0}^Nfrac{N^k}{k!}left(frac{n}{N}right)^kleqleft(frac{n}{N}right)^Ne^N$$
    at least once $n$ surpasses $N$. This inequality gives the limit we wanted.



    Now we are ready to start the formal presentation. Let $epsilon>0$. Choose $N_1$ in $mathbb{N}$ such that
    $$|a_n-a|<frac{epsilon}{2}$$
    for every $ngeq N_1$. Let $M$ satisfy the inequality below for every $n$ in $mathbb{N}$
    $$|a_n-a|<M,.$$
    Choose $N_2$ such that
    $$frac{1}{2^n}sum_{k=0}^{N_1-1}binom{n}{k}<frac{epsilon}{2(M+1)}$$
    for every $ngeq N_2$. We now observe the scratch-paper inequalities that inspired the approach:



    begin{align*}left|biggl(,sum_{k=0}^nfrac{1}{2^n}binom{n}{k}a_kbiggr)-a,right|
    &=left|sum_{k=0}^nfrac{1}{2^n}binom{n}{k}(a_k-a)right|\
    &leq frac{1}{2^n}sum_{k=0}^{N_1-1}binom{n}{k}|a_k-a|+sum_{k=N_1}^nfrac{1}{2^n}binom{n}{k}|a_k-a|
    end{align*}
    for sufficiently large $n$. In particular, if we take $n>max(N_1, N_2)$, we then get
    $$left|biggl(,sum_{k=0}^nfrac{1}{2^n}binom{n}{k}a_kbiggr)-a,right|<epsilon,.$$






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      Say $|a_n| le C$ for each $n$. For $epsilon > 0$, take $N$ s.t. $|a_n-a| < epsilon$ for $n ge N$. Then for $M ge N^2$, $$frac{1}{2^M}sum_{k=0}^M {M choose k}|a_k-a| le frac{1}{2^M}sum_{k=0}^sqrt{M} {M choose k}2C+frac{1}{2^M}sum_{k=sqrt{M}}^M {M choose k}epsilon.$$ So, $limsup_{M to infty} |frac{1}{2^M}sum_{k=0}^M {M choose k} a_k - a| le epsilon$. This holds for each $epsilon > 0$.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Say $|a_n| le C$ for each $n$. For $epsilon > 0$, take $N$ s.t. $|a_n-a| < epsilon$ for $n ge N$. Then for $M ge N^2$, $$frac{1}{2^M}sum_{k=0}^M {M choose k}|a_k-a| le frac{1}{2^M}sum_{k=0}^sqrt{M} {M choose k}2C+frac{1}{2^M}sum_{k=sqrt{M}}^M {M choose k}epsilon.$$ So, $limsup_{M to infty} |frac{1}{2^M}sum_{k=0}^M {M choose k} a_k - a| le epsilon$. This holds for each $epsilon > 0$.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Say $|a_n| le C$ for each $n$. For $epsilon > 0$, take $N$ s.t. $|a_n-a| < epsilon$ for $n ge N$. Then for $M ge N^2$, $$frac{1}{2^M}sum_{k=0}^M {M choose k}|a_k-a| le frac{1}{2^M}sum_{k=0}^sqrt{M} {M choose k}2C+frac{1}{2^M}sum_{k=sqrt{M}}^M {M choose k}epsilon.$$ So, $limsup_{M to infty} |frac{1}{2^M}sum_{k=0}^M {M choose k} a_k - a| le epsilon$. This holds for each $epsilon > 0$.






          share|cite|improve this answer









          $endgroup$



          Say $|a_n| le C$ for each $n$. For $epsilon > 0$, take $N$ s.t. $|a_n-a| < epsilon$ for $n ge N$. Then for $M ge N^2$, $$frac{1}{2^M}sum_{k=0}^M {M choose k}|a_k-a| le frac{1}{2^M}sum_{k=0}^sqrt{M} {M choose k}2C+frac{1}{2^M}sum_{k=sqrt{M}}^M {M choose k}epsilon.$$ So, $limsup_{M to infty} |frac{1}{2^M}sum_{k=0}^M {M choose k} a_k - a| le epsilon$. This holds for each $epsilon > 0$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 14 '18 at 3:41









          mathworker21mathworker21

          8,9421928




          8,9421928























              0












              $begingroup$

              There's a pretty standard exercise in which we are asked to show that if $a_krightarrow a$ then
              $$biggl(frac{1}{n}sum_{k=1}^n a_kbiggr)rightarrow a,.$$
              It's reasonable to suspect that a similar approach there might work here.



              After one starts down this road, you eventually realize that you'll need to show that




              $$lim_{nrightarrowinfty}frac{1}{2^n}sum_{k=0}^Nbinom{n}{k}=0$$




              where $N$ is some possibly-big, but-nevertheless-fixed positive integer. To accomplish this, we notice
              $$sum_{k=0}^Nbinom{n}{k}leqsum_{k=0}^Nfrac{n^k}{k!}=sum_{k=0}^Nfrac{N^k}{k!}left(frac{n}{N}right)^kleqleft(frac{n}{N}right)^Ne^N$$
              at least once $n$ surpasses $N$. This inequality gives the limit we wanted.



              Now we are ready to start the formal presentation. Let $epsilon>0$. Choose $N_1$ in $mathbb{N}$ such that
              $$|a_n-a|<frac{epsilon}{2}$$
              for every $ngeq N_1$. Let $M$ satisfy the inequality below for every $n$ in $mathbb{N}$
              $$|a_n-a|<M,.$$
              Choose $N_2$ such that
              $$frac{1}{2^n}sum_{k=0}^{N_1-1}binom{n}{k}<frac{epsilon}{2(M+1)}$$
              for every $ngeq N_2$. We now observe the scratch-paper inequalities that inspired the approach:



              begin{align*}left|biggl(,sum_{k=0}^nfrac{1}{2^n}binom{n}{k}a_kbiggr)-a,right|
              &=left|sum_{k=0}^nfrac{1}{2^n}binom{n}{k}(a_k-a)right|\
              &leq frac{1}{2^n}sum_{k=0}^{N_1-1}binom{n}{k}|a_k-a|+sum_{k=N_1}^nfrac{1}{2^n}binom{n}{k}|a_k-a|
              end{align*}
              for sufficiently large $n$. In particular, if we take $n>max(N_1, N_2)$, we then get
              $$left|biggl(,sum_{k=0}^nfrac{1}{2^n}binom{n}{k}a_kbiggr)-a,right|<epsilon,.$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                There's a pretty standard exercise in which we are asked to show that if $a_krightarrow a$ then
                $$biggl(frac{1}{n}sum_{k=1}^n a_kbiggr)rightarrow a,.$$
                It's reasonable to suspect that a similar approach there might work here.



                After one starts down this road, you eventually realize that you'll need to show that




                $$lim_{nrightarrowinfty}frac{1}{2^n}sum_{k=0}^Nbinom{n}{k}=0$$




                where $N$ is some possibly-big, but-nevertheless-fixed positive integer. To accomplish this, we notice
                $$sum_{k=0}^Nbinom{n}{k}leqsum_{k=0}^Nfrac{n^k}{k!}=sum_{k=0}^Nfrac{N^k}{k!}left(frac{n}{N}right)^kleqleft(frac{n}{N}right)^Ne^N$$
                at least once $n$ surpasses $N$. This inequality gives the limit we wanted.



                Now we are ready to start the formal presentation. Let $epsilon>0$. Choose $N_1$ in $mathbb{N}$ such that
                $$|a_n-a|<frac{epsilon}{2}$$
                for every $ngeq N_1$. Let $M$ satisfy the inequality below for every $n$ in $mathbb{N}$
                $$|a_n-a|<M,.$$
                Choose $N_2$ such that
                $$frac{1}{2^n}sum_{k=0}^{N_1-1}binom{n}{k}<frac{epsilon}{2(M+1)}$$
                for every $ngeq N_2$. We now observe the scratch-paper inequalities that inspired the approach:



                begin{align*}left|biggl(,sum_{k=0}^nfrac{1}{2^n}binom{n}{k}a_kbiggr)-a,right|
                &=left|sum_{k=0}^nfrac{1}{2^n}binom{n}{k}(a_k-a)right|\
                &leq frac{1}{2^n}sum_{k=0}^{N_1-1}binom{n}{k}|a_k-a|+sum_{k=N_1}^nfrac{1}{2^n}binom{n}{k}|a_k-a|
                end{align*}
                for sufficiently large $n$. In particular, if we take $n>max(N_1, N_2)$, we then get
                $$left|biggl(,sum_{k=0}^nfrac{1}{2^n}binom{n}{k}a_kbiggr)-a,right|<epsilon,.$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  There's a pretty standard exercise in which we are asked to show that if $a_krightarrow a$ then
                  $$biggl(frac{1}{n}sum_{k=1}^n a_kbiggr)rightarrow a,.$$
                  It's reasonable to suspect that a similar approach there might work here.



                  After one starts down this road, you eventually realize that you'll need to show that




                  $$lim_{nrightarrowinfty}frac{1}{2^n}sum_{k=0}^Nbinom{n}{k}=0$$




                  where $N$ is some possibly-big, but-nevertheless-fixed positive integer. To accomplish this, we notice
                  $$sum_{k=0}^Nbinom{n}{k}leqsum_{k=0}^Nfrac{n^k}{k!}=sum_{k=0}^Nfrac{N^k}{k!}left(frac{n}{N}right)^kleqleft(frac{n}{N}right)^Ne^N$$
                  at least once $n$ surpasses $N$. This inequality gives the limit we wanted.



                  Now we are ready to start the formal presentation. Let $epsilon>0$. Choose $N_1$ in $mathbb{N}$ such that
                  $$|a_n-a|<frac{epsilon}{2}$$
                  for every $ngeq N_1$. Let $M$ satisfy the inequality below for every $n$ in $mathbb{N}$
                  $$|a_n-a|<M,.$$
                  Choose $N_2$ such that
                  $$frac{1}{2^n}sum_{k=0}^{N_1-1}binom{n}{k}<frac{epsilon}{2(M+1)}$$
                  for every $ngeq N_2$. We now observe the scratch-paper inequalities that inspired the approach:



                  begin{align*}left|biggl(,sum_{k=0}^nfrac{1}{2^n}binom{n}{k}a_kbiggr)-a,right|
                  &=left|sum_{k=0}^nfrac{1}{2^n}binom{n}{k}(a_k-a)right|\
                  &leq frac{1}{2^n}sum_{k=0}^{N_1-1}binom{n}{k}|a_k-a|+sum_{k=N_1}^nfrac{1}{2^n}binom{n}{k}|a_k-a|
                  end{align*}
                  for sufficiently large $n$. In particular, if we take $n>max(N_1, N_2)$, we then get
                  $$left|biggl(,sum_{k=0}^nfrac{1}{2^n}binom{n}{k}a_kbiggr)-a,right|<epsilon,.$$






                  share|cite|improve this answer









                  $endgroup$



                  There's a pretty standard exercise in which we are asked to show that if $a_krightarrow a$ then
                  $$biggl(frac{1}{n}sum_{k=1}^n a_kbiggr)rightarrow a,.$$
                  It's reasonable to suspect that a similar approach there might work here.



                  After one starts down this road, you eventually realize that you'll need to show that




                  $$lim_{nrightarrowinfty}frac{1}{2^n}sum_{k=0}^Nbinom{n}{k}=0$$




                  where $N$ is some possibly-big, but-nevertheless-fixed positive integer. To accomplish this, we notice
                  $$sum_{k=0}^Nbinom{n}{k}leqsum_{k=0}^Nfrac{n^k}{k!}=sum_{k=0}^Nfrac{N^k}{k!}left(frac{n}{N}right)^kleqleft(frac{n}{N}right)^Ne^N$$
                  at least once $n$ surpasses $N$. This inequality gives the limit we wanted.



                  Now we are ready to start the formal presentation. Let $epsilon>0$. Choose $N_1$ in $mathbb{N}$ such that
                  $$|a_n-a|<frac{epsilon}{2}$$
                  for every $ngeq N_1$. Let $M$ satisfy the inequality below for every $n$ in $mathbb{N}$
                  $$|a_n-a|<M,.$$
                  Choose $N_2$ such that
                  $$frac{1}{2^n}sum_{k=0}^{N_1-1}binom{n}{k}<frac{epsilon}{2(M+1)}$$
                  for every $ngeq N_2$. We now observe the scratch-paper inequalities that inspired the approach:



                  begin{align*}left|biggl(,sum_{k=0}^nfrac{1}{2^n}binom{n}{k}a_kbiggr)-a,right|
                  &=left|sum_{k=0}^nfrac{1}{2^n}binom{n}{k}(a_k-a)right|\
                  &leq frac{1}{2^n}sum_{k=0}^{N_1-1}binom{n}{k}|a_k-a|+sum_{k=N_1}^nfrac{1}{2^n}binom{n}{k}|a_k-a|
                  end{align*}
                  for sufficiently large $n$. In particular, if we take $n>max(N_1, N_2)$, we then get
                  $$left|biggl(,sum_{k=0}^nfrac{1}{2^n}binom{n}{k}a_kbiggr)-a,right|<epsilon,.$$







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                  answered Dec 14 '18 at 14:35









                  Robert WolfeRobert Wolfe

                  5,79722763




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