Understanding part of derivation of Chebychev's Theorem

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I cannot understand this result from pages 17–18 of Tenenbaum and Mendes's The Prime Numbers and Their Distribution on how the summation of $fracxlog(2)2^j+O(log(x))$ results in $2x log(2) + O(log(x)^2).$
What are the steps in the summation that give the result? Did they use a telescoping series?



It is given that $B_2(x)=log[2x]! - 2log[x]! = B(x)-2B(x/2)$, and from the text,




The lower estimate for $B_2 (x)$ is used inductively: we have
beginalign psi(x) & le B_2(x) + psi(x/2) le B_2(x) + B_2(x/2) + psi(x/4) \
& le cdots le sum_0 le j le k B_2(x/2^j) + psi(x/2^k+1). endalign
Here, $k$ is an arbitrary integer. Let us choose $$k = K(x) := [(log x)/ log 2],$$
so that $psi(x/2^k+1) = 0$. It follows that
beginalign psi(x) & le sum_0 le j le K(x) leftlbrace fracx log 22^j + O(log x)rightrbrace \
& le 2x log 2 + O((log x)^2).
endalign





number-theory elementary-number-theory prime-numbers analytic-number-theory prime-factorization




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    I cannot understand this result from pages 17–18 of Tenenbaum and Mendes's The Prime Numbers and Their Distribution on how the summation of $fracxlog(2)2^j+O(log(x))$ results in $2x log(2) + O(log(x)^2).$
    What are the steps in the summation that give the result? Did they use a telescoping series?



    It is given that $B_2(x)=log[2x]! - 2log[x]! = B(x)-2B(x/2)$, and from the text,




    The lower estimate for $B_2 (x)$ is used inductively: we have
    beginalign psi(x) & le B_2(x) + psi(x/2) le B_2(x) + B_2(x/2) + psi(x/4) \
    & le cdots le sum_0 le j le k B_2(x/2^j) + psi(x/2^k+1). endalign
    Here, $k$ is an arbitrary integer. Let us choose $$k = K(x) := [(log x)/ log 2],$$
    so that $psi(x/2^k+1) = 0$. It follows that
    beginalign psi(x) & le sum_0 le j le K(x) leftlbrace fracx log 22^j + O(log x)rightrbrace \
    & le 2x log 2 + O((log x)^2).
    endalign





    number-theory elementary-number-theory prime-numbers analytic-number-theory prime-factorization




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      I cannot understand this result from pages 17–18 of Tenenbaum and Mendes's The Prime Numbers and Their Distribution on how the summation of $fracxlog(2)2^j+O(log(x))$ results in $2x log(2) + O(log(x)^2).$
      What are the steps in the summation that give the result? Did they use a telescoping series?



      It is given that $B_2(x)=log[2x]! - 2log[x]! = B(x)-2B(x/2)$, and from the text,




      The lower estimate for $B_2 (x)$ is used inductively: we have
      beginalign psi(x) & le B_2(x) + psi(x/2) le B_2(x) + B_2(x/2) + psi(x/4) \
      & le cdots le sum_0 le j le k B_2(x/2^j) + psi(x/2^k+1). endalign
      Here, $k$ is an arbitrary integer. Let us choose $$k = K(x) := [(log x)/ log 2],$$
      so that $psi(x/2^k+1) = 0$. It follows that
      beginalign psi(x) & le sum_0 le j le K(x) leftlbrace fracx log 22^j + O(log x)rightrbrace \
      & le 2x log 2 + O((log x)^2).
      endalign





      number-theory elementary-number-theory prime-numbers analytic-number-theory prime-factorization




      share|cite|improve this question













      I cannot understand this result from pages 17–18 of Tenenbaum and Mendes's The Prime Numbers and Their Distribution on how the summation of $fracxlog(2)2^j+O(log(x))$ results in $2x log(2) + O(log(x)^2).$
      What are the steps in the summation that give the result? Did they use a telescoping series?



      It is given that $B_2(x)=log[2x]! - 2log[x]! = B(x)-2B(x/2)$, and from the text,




      The lower estimate for $B_2 (x)$ is used inductively: we have
      beginalign psi(x) & le B_2(x) + psi(x/2) le B_2(x) + B_2(x/2) + psi(x/4) \
      & le cdots le sum_0 le j le k B_2(x/2^j) + psi(x/2^k+1). endalign
      Here, $k$ is an arbitrary integer. Let us choose $$k = K(x) := [(log x)/ log 2],$$
      so that $psi(x/2^k+1) = 0$. It follows that
      beginalign psi(x) & le sum_0 le j le K(x) leftlbrace fracx log 22^j + O(log x)rightrbrace \
      & le 2x log 2 + O((log x)^2).
      endalign






      number-theory elementary-number-theory prime-numbers analytic-number-theory prime-factorization





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      edited Aug 6 at 9:50
























      asked Aug 5 at 19:18









      onepound

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          $$sum_jleft(xlog2over2^j+O(log x)right)=xlog2sum_j(1/2^j)+sum_jO(log x)le2xlog2+O(K(x)(log x))$$



          $$=2xlog2+O((log x)^2)$$






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          • 1




            I see because $sum_j(1/2^j)=2$ and $log 2$ in asymptotic $((log x)/ log 2) log x$ can be ignored. Thanks.
            – onepound
            Aug 7 at 9:27











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          $$sum_jleft(xlog2over2^j+O(log x)right)=xlog2sum_j(1/2^j)+sum_jO(log x)le2xlog2+O(K(x)(log x))$$



          $$=2xlog2+O((log x)^2)$$






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          • 1




            I see because $sum_j(1/2^j)=2$ and $log 2$ in asymptotic $((log x)/ log 2) log x$ can be ignored. Thanks.
            – onepound
            Aug 7 at 9:27















          up vote
          1
          down vote



          accepted










          $$sum_jleft(xlog2over2^j+O(log x)right)=xlog2sum_j(1/2^j)+sum_jO(log x)le2xlog2+O(K(x)(log x))$$



          $$=2xlog2+O((log x)^2)$$






          share|cite|improve this answer

















          • 1




            I see because $sum_j(1/2^j)=2$ and $log 2$ in asymptotic $((log x)/ log 2) log x$ can be ignored. Thanks.
            – onepound
            Aug 7 at 9:27













          up vote
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          up vote
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          accepted






          $$sum_jleft(xlog2over2^j+O(log x)right)=xlog2sum_j(1/2^j)+sum_jO(log x)le2xlog2+O(K(x)(log x))$$



          $$=2xlog2+O((log x)^2)$$






          share|cite|improve this answer













          $$sum_jleft(xlog2over2^j+O(log x)right)=xlog2sum_j(1/2^j)+sum_jO(log x)le2xlog2+O(K(x)(log x))$$



          $$=2xlog2+O((log x)^2)$$







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          answered Aug 6 at 10:04









          Gerry Myerson

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          • 1




            I see because $sum_j(1/2^j)=2$ and $log 2$ in asymptotic $((log x)/ log 2) log x$ can be ignored. Thanks.
            – onepound
            Aug 7 at 9:27













          • 1




            I see because $sum_j(1/2^j)=2$ and $log 2$ in asymptotic $((log x)/ log 2) log x$ can be ignored. Thanks.
            – onepound
            Aug 7 at 9:27








          1




          1




          I see because $sum_j(1/2^j)=2$ and $log 2$ in asymptotic $((log x)/ log 2) log x$ can be ignored. Thanks.
          – onepound
          Aug 7 at 9:27





          I see because $sum_j(1/2^j)=2$ and $log 2$ in asymptotic $((log x)/ log 2) log x$ can be ignored. Thanks.
          – onepound
          Aug 7 at 9:27













           

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