Adventure with infinite series, a curiosity

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It is easily verifiable that
$$sum_kgeq0binom2kkfrac12^3k=sqrt2.$$
It is not that difficult to get
$$sum_kgeq0binom4k2kfrac12^5k=fracsqrt2-sqrt2+sqrt2+sqrt22.$$




Question. Is there something similarly "nice" in computing
$$sum_kgeq0binom8k4kfrac12^10k=?$$
Perhaps the same question about
$$sum_kgeq0binom16k8kfrac12^20k=?$$




NOTE. The powers of $2$ are selected with a hope (suspicion) for some pattern.






real-analysis sequences-and-series







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    It is easily verifiable that
    $$sum_kgeq0binom2kkfrac12^3k=sqrt2.$$
    It is not that difficult to get
    $$sum_kgeq0binom4k2kfrac12^5k=fracsqrt2-sqrt2+sqrt2+sqrt22.$$




    Question. Is there something similarly "nice" in computing
    $$sum_kgeq0binom8k4kfrac12^10k=?$$
    Perhaps the same question about
    $$sum_kgeq0binom16k8kfrac12^20k=?$$




    NOTE. The powers of $2$ are selected with a hope (suspicion) for some pattern.






    real-analysis sequences-and-series







    share|cite|improve this question























      up vote
      1
      down vote

      favorite
      1









      up vote
      1
      down vote

      favorite
      1






      1





      It is easily verifiable that
      $$sum_kgeq0binom2kkfrac12^3k=sqrt2.$$
      It is not that difficult to get
      $$sum_kgeq0binom4k2kfrac12^5k=fracsqrt2-sqrt2+sqrt2+sqrt22.$$




      Question. Is there something similarly "nice" in computing
      $$sum_kgeq0binom8k4kfrac12^10k=?$$
      Perhaps the same question about
      $$sum_kgeq0binom16k8kfrac12^20k=?$$




      NOTE. The powers of $2$ are selected with a hope (suspicion) for some pattern.






      real-analysis sequences-and-series







      share|cite|improve this question













      It is easily verifiable that
      $$sum_kgeq0binom2kkfrac12^3k=sqrt2.$$
      It is not that difficult to get
      $$sum_kgeq0binom4k2kfrac12^5k=fracsqrt2-sqrt2+sqrt2+sqrt22.$$




      Question. Is there something similarly "nice" in computing
      $$sum_kgeq0binom8k4kfrac12^10k=?$$
      Perhaps the same question about
      $$sum_kgeq0binom16k8kfrac12^20k=?$$




      NOTE. The powers of $2$ are selected with a hope (suspicion) for some pattern.






      real-analysis sequences-and-series




      real-analysis sequences-and-series






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      asked 1 hour ago









      T. Amdeberhan

      16.2k228120




      16.2k228120




















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          It is moderately nice, I would say. We have $sum binom2kk x^k=(1-4x)^-1/2$ for $|x|<1/4$. If we need only terms with $k$ divisible by 4 and $x=2^-5/2$, $4x=2^-1/2$, we get $$sum binom8k4k2^-10k=frac14sum_w^4=1(1-w/sqrt2)^-1/2$$
          and so on.






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            More generally, it seems that
            $$
            sum_kge0binom2^j+1k2^jk2^-a_j+2 k=2^-jsum_w^2^j=1(1-w/sqrt2)^-1/2
            $$
            where $a_1=2,a_2=3,a_j+1=a_j+a_j-1+dots+a_1$ (cf. A257113).






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            2 Answers
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            2 Answers
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            It is moderately nice, I would say. We have $sum binom2kk x^k=(1-4x)^-1/2$ for $|x|<1/4$. If we need only terms with $k$ divisible by 4 and $x=2^-5/2$, $4x=2^-1/2$, we get $$sum binom8k4k2^-10k=frac14sum_w^4=1(1-w/sqrt2)^-1/2$$
            and so on.






            share|cite|improve this answer
























              up vote
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              It is moderately nice, I would say. We have $sum binom2kk x^k=(1-4x)^-1/2$ for $|x|<1/4$. If we need only terms with $k$ divisible by 4 and $x=2^-5/2$, $4x=2^-1/2$, we get $$sum binom8k4k2^-10k=frac14sum_w^4=1(1-w/sqrt2)^-1/2$$
              and so on.






              share|cite|improve this answer






















                up vote
                5
                down vote










                up vote
                5
                down vote









                It is moderately nice, I would say. We have $sum binom2kk x^k=(1-4x)^-1/2$ for $|x|<1/4$. If we need only terms with $k$ divisible by 4 and $x=2^-5/2$, $4x=2^-1/2$, we get $$sum binom8k4k2^-10k=frac14sum_w^4=1(1-w/sqrt2)^-1/2$$
                and so on.






                share|cite|improve this answer












                It is moderately nice, I would say. We have $sum binom2kk x^k=(1-4x)^-1/2$ for $|x|<1/4$. If we need only terms with $k$ divisible by 4 and $x=2^-5/2$, $4x=2^-1/2$, we get $$sum binom8k4k2^-10k=frac14sum_w^4=1(1-w/sqrt2)^-1/2$$
                and so on.







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                share|cite|improve this answer



                share|cite|improve this answer










                answered 1 hour ago









                Fedor Petrov

                45.3k5108213




                45.3k5108213




















                    up vote
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                    More generally, it seems that
                    $$
                    sum_kge0binom2^j+1k2^jk2^-a_j+2 k=2^-jsum_w^2^j=1(1-w/sqrt2)^-1/2
                    $$
                    where $a_1=2,a_2=3,a_j+1=a_j+a_j-1+dots+a_1$ (cf. A257113).






                    share|cite|improve this answer




















                    • Note: this has been confirmed up to $j=7$.
                      – AccidentalFourierTransform
                      4 mins ago














                    up vote
                    0
                    down vote













                    More generally, it seems that
                    $$
                    sum_kge0binom2^j+1k2^jk2^-a_j+2 k=2^-jsum_w^2^j=1(1-w/sqrt2)^-1/2
                    $$
                    where $a_1=2,a_2=3,a_j+1=a_j+a_j-1+dots+a_1$ (cf. A257113).






                    share|cite|improve this answer




















                    • Note: this has been confirmed up to $j=7$.
                      – AccidentalFourierTransform
                      4 mins ago












                    up vote
                    0
                    down vote










                    up vote
                    0
                    down vote









                    More generally, it seems that
                    $$
                    sum_kge0binom2^j+1k2^jk2^-a_j+2 k=2^-jsum_w^2^j=1(1-w/sqrt2)^-1/2
                    $$
                    where $a_1=2,a_2=3,a_j+1=a_j+a_j-1+dots+a_1$ (cf. A257113).






                    share|cite|improve this answer












                    More generally, it seems that
                    $$
                    sum_kge0binom2^j+1k2^jk2^-a_j+2 k=2^-jsum_w^2^j=1(1-w/sqrt2)^-1/2
                    $$
                    where $a_1=2,a_2=3,a_j+1=a_j+a_j-1+dots+a_1$ (cf. A257113).







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 21 mins ago









                    AccidentalFourierTransform

                    541112




                    541112











                    • Note: this has been confirmed up to $j=7$.
                      – AccidentalFourierTransform
                      4 mins ago
















                    • Note: this has been confirmed up to $j=7$.
                      – AccidentalFourierTransform
                      4 mins ago















                    Note: this has been confirmed up to $j=7$.
                    – AccidentalFourierTransform
                    4 mins ago




                    Note: this has been confirmed up to $j=7$.
                    – AccidentalFourierTransform
                    4 mins ago

















                     

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