sum of squares of digits of number n












1












$begingroup$


I have to found a limit of $B/n$, where $B$ is a sum of squares of digits of number $n$. I was thinking about using Cesaro-Stolz theorem, but I'm stuck with the $B$.






analysis elementary-number-theory limits-without-lhopital







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Hint: $Bleq9^2log_{10}(n)$, more accurately $Bleq9^2(lfloorlog_{10}(n)rfloor+1)$
    $endgroup$
    – SmileyCraft
    Jan 10 at 19:45












  • $begingroup$
    Wow, I didn't know that fact. Thank you!
    $endgroup$
    – MilO
    Jan 10 at 19:46








  • 1




    $begingroup$
    You should try to justify it though
    $endgroup$
    – SmileyCraft
    Jan 10 at 19:47










  • $begingroup$
    Hint: suppose $n = underbrace{999cdots 999}_n$.
    $endgroup$
    – David G. Stork
    Jan 10 at 19:49


















1












$begingroup$


I have to found a limit of $B/n$, where $B$ is a sum of squares of digits of number $n$. I was thinking about using Cesaro-Stolz theorem, but I'm stuck with the $B$.






analysis elementary-number-theory limits-without-lhopital







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Hint: $Bleq9^2log_{10}(n)$, more accurately $Bleq9^2(lfloorlog_{10}(n)rfloor+1)$
    $endgroup$
    – SmileyCraft
    Jan 10 at 19:45












  • $begingroup$
    Wow, I didn't know that fact. Thank you!
    $endgroup$
    – MilO
    Jan 10 at 19:46








  • 1




    $begingroup$
    You should try to justify it though
    $endgroup$
    – SmileyCraft
    Jan 10 at 19:47










  • $begingroup$
    Hint: suppose $n = underbrace{999cdots 999}_n$.
    $endgroup$
    – David G. Stork
    Jan 10 at 19:49
















1












1








1





$begingroup$


I have to found a limit of $B/n$, where $B$ is a sum of squares of digits of number $n$. I was thinking about using Cesaro-Stolz theorem, but I'm stuck with the $B$.






analysis elementary-number-theory limits-without-lhopital







share|cite|improve this question











$endgroup$




I have to found a limit of $B/n$, where $B$ is a sum of squares of digits of number $n$. I was thinking about using Cesaro-Stolz theorem, but I'm stuck with the $B$.






analysis elementary-number-theory limits-without-lhopital




analysis elementary-number-theory limits-without-lhopital






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 10 at 20:03









David G. Stork

11k41432




11k41432










asked Jan 10 at 19:44









MilOMilO

83




83








  • 1




    $begingroup$
    Hint: $Bleq9^2log_{10}(n)$, more accurately $Bleq9^2(lfloorlog_{10}(n)rfloor+1)$
    $endgroup$
    – SmileyCraft
    Jan 10 at 19:45












  • $begingroup$
    Wow, I didn't know that fact. Thank you!
    $endgroup$
    – MilO
    Jan 10 at 19:46








  • 1




    $begingroup$
    You should try to justify it though
    $endgroup$
    – SmileyCraft
    Jan 10 at 19:47










  • $begingroup$
    Hint: suppose $n = underbrace{999cdots 999}_n$.
    $endgroup$
    – David G. Stork
    Jan 10 at 19:49
















  • 1




    $begingroup$
    Hint: $Bleq9^2log_{10}(n)$, more accurately $Bleq9^2(lfloorlog_{10}(n)rfloor+1)$
    $endgroup$
    – SmileyCraft
    Jan 10 at 19:45












  • $begingroup$
    Wow, I didn't know that fact. Thank you!
    $endgroup$
    – MilO
    Jan 10 at 19:46








  • 1




    $begingroup$
    You should try to justify it though
    $endgroup$
    – SmileyCraft
    Jan 10 at 19:47










  • $begingroup$
    Hint: suppose $n = underbrace{999cdots 999}_n$.
    $endgroup$
    – David G. Stork
    Jan 10 at 19:49










1




1




$begingroup$
Hint: $Bleq9^2log_{10}(n)$, more accurately $Bleq9^2(lfloorlog_{10}(n)rfloor+1)$
$endgroup$
– SmileyCraft
Jan 10 at 19:45






$begingroup$
Hint: $Bleq9^2log_{10}(n)$, more accurately $Bleq9^2(lfloorlog_{10}(n)rfloor+1)$
$endgroup$
– SmileyCraft
Jan 10 at 19:45














$begingroup$
Wow, I didn't know that fact. Thank you!
$endgroup$
– MilO
Jan 10 at 19:46






$begingroup$
Wow, I didn't know that fact. Thank you!
$endgroup$
– MilO
Jan 10 at 19:46






1




1




$begingroup$
You should try to justify it though
$endgroup$
– SmileyCraft
Jan 10 at 19:47




$begingroup$
You should try to justify it though
$endgroup$
– SmileyCraft
Jan 10 at 19:47












$begingroup$
Hint: suppose $n = underbrace{999cdots 999}_n$.
$endgroup$
– David G. Stork
Jan 10 at 19:49






$begingroup$
Hint: suppose $n = underbrace{999cdots 999}_n$.
$endgroup$
– David G. Stork
Jan 10 at 19:49












2 Answers
2






active

oldest

votes


















0












$begingroup$

Let $$n=overline{a_k a_{k-1}cdots a_0}=10^ka_k+cdots +a_0$$when $a_kne 0$. Therefore$${B_nover n}={a_0^2+cdots +a_k^2over 10^ka_k+cdots +a_0}le {81(k+1)over 10^k}$$and we obtain$$0le lim_{nto infty}{B_nover n}=lim_{kto infty}{a_0^2+cdots +a_k^2over 10^ka_k+cdots +a_0}le lim_{kto infty}{81(k+1)over 10^k}=0$$therefore$$lim_{nto infty}{B_nover n}=0$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Consider a number $n$
    in base $b$
    with $d$ digits,
    the high order digit
    being non-zero.



    Then
    $b^{d-1} le n < b^d$
    (so that
    $d-1 le log_b(n) < d$)
    and
    $B_m(n)$,
    the sum of the $m$-th powers of the digits of $n$
    in base $b$,
    satisfies



    $begin{array}\
    B_m(n)
    &le d(b-1)^m\
    &lt (log_b(n)+1)(b-1)^m\
    text{so}\
    dfrac{B_m(n)}{n}
    &lt dfrac{(log_b(n)+1)(b-1)^m}{n}\
    &= dfrac{(ln(n)+ln(b))(b-1)^m}{nln(b)}\
    &lt dfrac{(2sqrt{n}+ln(b))(b-1)^m}{nln(b)}
    qquadtext{since }ln(n) < 2sqrt{n}\
    &lt dfrac{3sqrt{n}(b-1)^m}{nln(b)}
    qquadtext{for } n > (ln(b))^2\
    &lt dfrac{3(b-1)^m}{sqrt{n}ln(b)}\
    &lt epsilon
    qquadtext{for } n > left(dfrac{3(b-1)^m}{epsilonln(b)}right)^2\
    &to 0\
    end{array}
    $



    This requires
    $n > dfrac{c(b, m)}{epsilon^2}
    $
    where $c(b, m)$
    is an expression that depends on
    $b$ and $m$.



    For any $a > 0$,
    this can be improved to
    $n > dfrac{c(b, m, a)}{epsilon^{1+a}}
    $
    where $c(b, m)$
    is a constant that depends on
    $b$, $m$, and $a$.



    I don't know if
    this can be improved to
    $n > dfrac{c_0(b, m)}{epsilon}
    $
    for some $c_0(b, m)$.






    share|cite|improve this answer









    $endgroup$













      Your Answer





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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      Let $$n=overline{a_k a_{k-1}cdots a_0}=10^ka_k+cdots +a_0$$when $a_kne 0$. Therefore$${B_nover n}={a_0^2+cdots +a_k^2over 10^ka_k+cdots +a_0}le {81(k+1)over 10^k}$$and we obtain$$0le lim_{nto infty}{B_nover n}=lim_{kto infty}{a_0^2+cdots +a_k^2over 10^ka_k+cdots +a_0}le lim_{kto infty}{81(k+1)over 10^k}=0$$therefore$$lim_{nto infty}{B_nover n}=0$$






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        Let $$n=overline{a_k a_{k-1}cdots a_0}=10^ka_k+cdots +a_0$$when $a_kne 0$. Therefore$${B_nover n}={a_0^2+cdots +a_k^2over 10^ka_k+cdots +a_0}le {81(k+1)over 10^k}$$and we obtain$$0le lim_{nto infty}{B_nover n}=lim_{kto infty}{a_0^2+cdots +a_k^2over 10^ka_k+cdots +a_0}le lim_{kto infty}{81(k+1)over 10^k}=0$$therefore$$lim_{nto infty}{B_nover n}=0$$






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          Let $$n=overline{a_k a_{k-1}cdots a_0}=10^ka_k+cdots +a_0$$when $a_kne 0$. Therefore$${B_nover n}={a_0^2+cdots +a_k^2over 10^ka_k+cdots +a_0}le {81(k+1)over 10^k}$$and we obtain$$0le lim_{nto infty}{B_nover n}=lim_{kto infty}{a_0^2+cdots +a_k^2over 10^ka_k+cdots +a_0}le lim_{kto infty}{81(k+1)over 10^k}=0$$therefore$$lim_{nto infty}{B_nover n}=0$$






          share|cite|improve this answer









          $endgroup$



          Let $$n=overline{a_k a_{k-1}cdots a_0}=10^ka_k+cdots +a_0$$when $a_kne 0$. Therefore$${B_nover n}={a_0^2+cdots +a_k^2over 10^ka_k+cdots +a_0}le {81(k+1)over 10^k}$$and we obtain$$0le lim_{nto infty}{B_nover n}=lim_{kto infty}{a_0^2+cdots +a_k^2over 10^ka_k+cdots +a_0}le lim_{kto infty}{81(k+1)over 10^k}=0$$therefore$$lim_{nto infty}{B_nover n}=0$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 10 at 19:56









          Mostafa AyazMostafa Ayaz

          15.7k3939




          15.7k3939























              1












              $begingroup$

              Consider a number $n$
              in base $b$
              with $d$ digits,
              the high order digit
              being non-zero.



              Then
              $b^{d-1} le n < b^d$
              (so that
              $d-1 le log_b(n) < d$)
              and
              $B_m(n)$,
              the sum of the $m$-th powers of the digits of $n$
              in base $b$,
              satisfies



              $begin{array}\
              B_m(n)
              &le d(b-1)^m\
              &lt (log_b(n)+1)(b-1)^m\
              text{so}\
              dfrac{B_m(n)}{n}
              &lt dfrac{(log_b(n)+1)(b-1)^m}{n}\
              &= dfrac{(ln(n)+ln(b))(b-1)^m}{nln(b)}\
              &lt dfrac{(2sqrt{n}+ln(b))(b-1)^m}{nln(b)}
              qquadtext{since }ln(n) < 2sqrt{n}\
              &lt dfrac{3sqrt{n}(b-1)^m}{nln(b)}
              qquadtext{for } n > (ln(b))^2\
              &lt dfrac{3(b-1)^m}{sqrt{n}ln(b)}\
              &lt epsilon
              qquadtext{for } n > left(dfrac{3(b-1)^m}{epsilonln(b)}right)^2\
              &to 0\
              end{array}
              $



              This requires
              $n > dfrac{c(b, m)}{epsilon^2}
              $
              where $c(b, m)$
              is an expression that depends on
              $b$ and $m$.



              For any $a > 0$,
              this can be improved to
              $n > dfrac{c(b, m, a)}{epsilon^{1+a}}
              $
              where $c(b, m)$
              is a constant that depends on
              $b$, $m$, and $a$.



              I don't know if
              this can be improved to
              $n > dfrac{c_0(b, m)}{epsilon}
              $
              for some $c_0(b, m)$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Consider a number $n$
                in base $b$
                with $d$ digits,
                the high order digit
                being non-zero.



                Then
                $b^{d-1} le n < b^d$
                (so that
                $d-1 le log_b(n) < d$)
                and
                $B_m(n)$,
                the sum of the $m$-th powers of the digits of $n$
                in base $b$,
                satisfies



                $begin{array}\
                B_m(n)
                &le d(b-1)^m\
                &lt (log_b(n)+1)(b-1)^m\
                text{so}\
                dfrac{B_m(n)}{n}
                &lt dfrac{(log_b(n)+1)(b-1)^m}{n}\
                &= dfrac{(ln(n)+ln(b))(b-1)^m}{nln(b)}\
                &lt dfrac{(2sqrt{n}+ln(b))(b-1)^m}{nln(b)}
                qquadtext{since }ln(n) < 2sqrt{n}\
                &lt dfrac{3sqrt{n}(b-1)^m}{nln(b)}
                qquadtext{for } n > (ln(b))^2\
                &lt dfrac{3(b-1)^m}{sqrt{n}ln(b)}\
                &lt epsilon
                qquadtext{for } n > left(dfrac{3(b-1)^m}{epsilonln(b)}right)^2\
                &to 0\
                end{array}
                $



                This requires
                $n > dfrac{c(b, m)}{epsilon^2}
                $
                where $c(b, m)$
                is an expression that depends on
                $b$ and $m$.



                For any $a > 0$,
                this can be improved to
                $n > dfrac{c(b, m, a)}{epsilon^{1+a}}
                $
                where $c(b, m)$
                is a constant that depends on
                $b$, $m$, and $a$.



                I don't know if
                this can be improved to
                $n > dfrac{c_0(b, m)}{epsilon}
                $
                for some $c_0(b, m)$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Consider a number $n$
                  in base $b$
                  with $d$ digits,
                  the high order digit
                  being non-zero.



                  Then
                  $b^{d-1} le n < b^d$
                  (so that
                  $d-1 le log_b(n) < d$)
                  and
                  $B_m(n)$,
                  the sum of the $m$-th powers of the digits of $n$
                  in base $b$,
                  satisfies



                  $begin{array}\
                  B_m(n)
                  &le d(b-1)^m\
                  &lt (log_b(n)+1)(b-1)^m\
                  text{so}\
                  dfrac{B_m(n)}{n}
                  &lt dfrac{(log_b(n)+1)(b-1)^m}{n}\
                  &= dfrac{(ln(n)+ln(b))(b-1)^m}{nln(b)}\
                  &lt dfrac{(2sqrt{n}+ln(b))(b-1)^m}{nln(b)}
                  qquadtext{since }ln(n) < 2sqrt{n}\
                  &lt dfrac{3sqrt{n}(b-1)^m}{nln(b)}
                  qquadtext{for } n > (ln(b))^2\
                  &lt dfrac{3(b-1)^m}{sqrt{n}ln(b)}\
                  &lt epsilon
                  qquadtext{for } n > left(dfrac{3(b-1)^m}{epsilonln(b)}right)^2\
                  &to 0\
                  end{array}
                  $



                  This requires
                  $n > dfrac{c(b, m)}{epsilon^2}
                  $
                  where $c(b, m)$
                  is an expression that depends on
                  $b$ and $m$.



                  For any $a > 0$,
                  this can be improved to
                  $n > dfrac{c(b, m, a)}{epsilon^{1+a}}
                  $
                  where $c(b, m)$
                  is a constant that depends on
                  $b$, $m$, and $a$.



                  I don't know if
                  this can be improved to
                  $n > dfrac{c_0(b, m)}{epsilon}
                  $
                  for some $c_0(b, m)$.






                  share|cite|improve this answer









                  $endgroup$



                  Consider a number $n$
                  in base $b$
                  with $d$ digits,
                  the high order digit
                  being non-zero.



                  Then
                  $b^{d-1} le n < b^d$
                  (so that
                  $d-1 le log_b(n) < d$)
                  and
                  $B_m(n)$,
                  the sum of the $m$-th powers of the digits of $n$
                  in base $b$,
                  satisfies



                  $begin{array}\
                  B_m(n)
                  &le d(b-1)^m\
                  &lt (log_b(n)+1)(b-1)^m\
                  text{so}\
                  dfrac{B_m(n)}{n}
                  &lt dfrac{(log_b(n)+1)(b-1)^m}{n}\
                  &= dfrac{(ln(n)+ln(b))(b-1)^m}{nln(b)}\
                  &lt dfrac{(2sqrt{n}+ln(b))(b-1)^m}{nln(b)}
                  qquadtext{since }ln(n) < 2sqrt{n}\
                  &lt dfrac{3sqrt{n}(b-1)^m}{nln(b)}
                  qquadtext{for } n > (ln(b))^2\
                  &lt dfrac{3(b-1)^m}{sqrt{n}ln(b)}\
                  &lt epsilon
                  qquadtext{for } n > left(dfrac{3(b-1)^m}{epsilonln(b)}right)^2\
                  &to 0\
                  end{array}
                  $



                  This requires
                  $n > dfrac{c(b, m)}{epsilon^2}
                  $
                  where $c(b, m)$
                  is an expression that depends on
                  $b$ and $m$.



                  For any $a > 0$,
                  this can be improved to
                  $n > dfrac{c(b, m, a)}{epsilon^{1+a}}
                  $
                  where $c(b, m)$
                  is a constant that depends on
                  $b$, $m$, and $a$.



                  I don't know if
                  this can be improved to
                  $n > dfrac{c_0(b, m)}{epsilon}
                  $
                  for some $c_0(b, m)$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 10 at 21:52









                  marty cohenmarty cohen

                  74.2k549128




                  74.2k549128






























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