Rational as series?












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I was checking out a few things in the geometric series and realized all rational numbers can be shown as a geometric series.I was pretty sure I read something like that somewhere.Can anyone tell me which mathematician is it shown by?ps:- the formula i discovered is that a rational number
p/q=p/(q+1)+{p/(q+1)^2}+{p/(q+1)^3}+...






sequences-and-series reference-request rational-numbers geometric-series







share|cite|improve this question











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  • $begingroup$
    Any number $N$, rational or irrational, can be expressed as the sum of a convergent geometric series: $$N=frac N2+frac N4+frac N8+cdots$$
    $endgroup$
    – bof
    Feb 2 at 5:45










  • $begingroup$
    Follow-up to bof's comment. All you need is the existence of a series $sum a_n$ that converges to some number $a$. Then, given a number $N$, you can make a new series $sum b_n$ with $b_n = N a_n / a$ so that $sum b_n = sum N a_n / a = N / a sum a_n = N$. In their comment above, bof used the series defined by $a_n = 2^{-n}$ which converges to $a = 1$.
    $endgroup$
    – parsiad
    Feb 2 at 5:50












  • $begingroup$
    When proving a repeating decimal must be a rational number, we will use geometric series.
    $endgroup$
    – Eric Yau
    Feb 2 at 5:55










  • $begingroup$
    Thanks a lot man.....but can anyone tell me which rule does the series in my question follow?
    $endgroup$
    – Mahiyan Intesir Prethu
    Feb 2 at 6:03
















1












$begingroup$


I was checking out a few things in the geometric series and realized all rational numbers can be shown as a geometric series.I was pretty sure I read something like that somewhere.Can anyone tell me which mathematician is it shown by?ps:- the formula i discovered is that a rational number
p/q=p/(q+1)+{p/(q+1)^2}+{p/(q+1)^3}+...






sequences-and-series reference-request rational-numbers geometric-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    Any number $N$, rational or irrational, can be expressed as the sum of a convergent geometric series: $$N=frac N2+frac N4+frac N8+cdots$$
    $endgroup$
    – bof
    Feb 2 at 5:45










  • $begingroup$
    Follow-up to bof's comment. All you need is the existence of a series $sum a_n$ that converges to some number $a$. Then, given a number $N$, you can make a new series $sum b_n$ with $b_n = N a_n / a$ so that $sum b_n = sum N a_n / a = N / a sum a_n = N$. In their comment above, bof used the series defined by $a_n = 2^{-n}$ which converges to $a = 1$.
    $endgroup$
    – parsiad
    Feb 2 at 5:50












  • $begingroup$
    When proving a repeating decimal must be a rational number, we will use geometric series.
    $endgroup$
    – Eric Yau
    Feb 2 at 5:55










  • $begingroup$
    Thanks a lot man.....but can anyone tell me which rule does the series in my question follow?
    $endgroup$
    – Mahiyan Intesir Prethu
    Feb 2 at 6:03














1












1








1





$begingroup$


I was checking out a few things in the geometric series and realized all rational numbers can be shown as a geometric series.I was pretty sure I read something like that somewhere.Can anyone tell me which mathematician is it shown by?ps:- the formula i discovered is that a rational number
p/q=p/(q+1)+{p/(q+1)^2}+{p/(q+1)^3}+...






sequences-and-series reference-request rational-numbers geometric-series







share|cite|improve this question











$endgroup$




I was checking out a few things in the geometric series and realized all rational numbers can be shown as a geometric series.I was pretty sure I read something like that somewhere.Can anyone tell me which mathematician is it shown by?ps:- the formula i discovered is that a rational number
p/q=p/(q+1)+{p/(q+1)^2}+{p/(q+1)^3}+...






sequences-and-series reference-request rational-numbers geometric-series




sequences-and-series reference-request rational-numbers geometric-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 2 at 6:35







Mahiyan Intesir Prethu

















asked Feb 2 at 5:40









Mahiyan Intesir PrethuMahiyan Intesir Prethu

64




64












  • $begingroup$
    Any number $N$, rational or irrational, can be expressed as the sum of a convergent geometric series: $$N=frac N2+frac N4+frac N8+cdots$$
    $endgroup$
    – bof
    Feb 2 at 5:45










  • $begingroup$
    Follow-up to bof's comment. All you need is the existence of a series $sum a_n$ that converges to some number $a$. Then, given a number $N$, you can make a new series $sum b_n$ with $b_n = N a_n / a$ so that $sum b_n = sum N a_n / a = N / a sum a_n = N$. In their comment above, bof used the series defined by $a_n = 2^{-n}$ which converges to $a = 1$.
    $endgroup$
    – parsiad
    Feb 2 at 5:50












  • $begingroup$
    When proving a repeating decimal must be a rational number, we will use geometric series.
    $endgroup$
    – Eric Yau
    Feb 2 at 5:55










  • $begingroup$
    Thanks a lot man.....but can anyone tell me which rule does the series in my question follow?
    $endgroup$
    – Mahiyan Intesir Prethu
    Feb 2 at 6:03


















  • $begingroup$
    Any number $N$, rational or irrational, can be expressed as the sum of a convergent geometric series: $$N=frac N2+frac N4+frac N8+cdots$$
    $endgroup$
    – bof
    Feb 2 at 5:45










  • $begingroup$
    Follow-up to bof's comment. All you need is the existence of a series $sum a_n$ that converges to some number $a$. Then, given a number $N$, you can make a new series $sum b_n$ with $b_n = N a_n / a$ so that $sum b_n = sum N a_n / a = N / a sum a_n = N$. In their comment above, bof used the series defined by $a_n = 2^{-n}$ which converges to $a = 1$.
    $endgroup$
    – parsiad
    Feb 2 at 5:50












  • $begingroup$
    When proving a repeating decimal must be a rational number, we will use geometric series.
    $endgroup$
    – Eric Yau
    Feb 2 at 5:55










  • $begingroup$
    Thanks a lot man.....but can anyone tell me which rule does the series in my question follow?
    $endgroup$
    – Mahiyan Intesir Prethu
    Feb 2 at 6:03
















$begingroup$
Any number $N$, rational or irrational, can be expressed as the sum of a convergent geometric series: $$N=frac N2+frac N4+frac N8+cdots$$
$endgroup$
– bof
Feb 2 at 5:45




$begingroup$
Any number $N$, rational or irrational, can be expressed as the sum of a convergent geometric series: $$N=frac N2+frac N4+frac N8+cdots$$
$endgroup$
– bof
Feb 2 at 5:45












$begingroup$
Follow-up to bof's comment. All you need is the existence of a series $sum a_n$ that converges to some number $a$. Then, given a number $N$, you can make a new series $sum b_n$ with $b_n = N a_n / a$ so that $sum b_n = sum N a_n / a = N / a sum a_n = N$. In their comment above, bof used the series defined by $a_n = 2^{-n}$ which converges to $a = 1$.
$endgroup$
– parsiad
Feb 2 at 5:50






$begingroup$
Follow-up to bof's comment. All you need is the existence of a series $sum a_n$ that converges to some number $a$. Then, given a number $N$, you can make a new series $sum b_n$ with $b_n = N a_n / a$ so that $sum b_n = sum N a_n / a = N / a sum a_n = N$. In their comment above, bof used the series defined by $a_n = 2^{-n}$ which converges to $a = 1$.
$endgroup$
– parsiad
Feb 2 at 5:50














$begingroup$
When proving a repeating decimal must be a rational number, we will use geometric series.
$endgroup$
– Eric Yau
Feb 2 at 5:55




$begingroup$
When proving a repeating decimal must be a rational number, we will use geometric series.
$endgroup$
– Eric Yau
Feb 2 at 5:55












$begingroup$
Thanks a lot man.....but can anyone tell me which rule does the series in my question follow?
$endgroup$
– Mahiyan Intesir Prethu
Feb 2 at 6:03




$begingroup$
Thanks a lot man.....but can anyone tell me which rule does the series in my question follow?
$endgroup$
– Mahiyan Intesir Prethu
Feb 2 at 6:03










2 Answers
2






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oldest

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1












$begingroup$

You have
$$
sum_{n=1}^infty frac{1}{(q+1)^n} = frac{1}{q+1} cdot frac{1}{1-frac{1}{q+1}}= frac{1}{q}
$$
using the formula for geometric series applied to $r = frac{1}{q+1}$, recalling that $0< frac{1}{q+1} < 1$.
Therefore,
$$
sum_{n=1}^infty frac{p}{(q+1)^n}= frac{p}{q}
$$
as claimed.



I don't know who showed this first, but this is pretty straightforward.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    The formula you wrote is not entirely correct. What you have is
    $$tag1
    frac pq=sum_{k=1}^infty left(frac p{p+q}right)^k.
    $$
    This works nicely when $p,q>0$, since in this case $left|tfrac p{p+q}right|<1$. The formula does not work for many negative numbers.



    As for who may have come up with such formula, the geometric series has been known for millennia: the formula, at least for a finite sum, already appears in Euclid's work. This is basic enough that I doubt it ever appeared as a theorem in anybody's work.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

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      active

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      1












      $begingroup$

      You have
      $$
      sum_{n=1}^infty frac{1}{(q+1)^n} = frac{1}{q+1} cdot frac{1}{1-frac{1}{q+1}}= frac{1}{q}
      $$
      using the formula for geometric series applied to $r = frac{1}{q+1}$, recalling that $0< frac{1}{q+1} < 1$.
      Therefore,
      $$
      sum_{n=1}^infty frac{p}{(q+1)^n}= frac{p}{q}
      $$
      as claimed.



      I don't know who showed this first, but this is pretty straightforward.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        You have
        $$
        sum_{n=1}^infty frac{1}{(q+1)^n} = frac{1}{q+1} cdot frac{1}{1-frac{1}{q+1}}= frac{1}{q}
        $$
        using the formula for geometric series applied to $r = frac{1}{q+1}$, recalling that $0< frac{1}{q+1} < 1$.
        Therefore,
        $$
        sum_{n=1}^infty frac{p}{(q+1)^n}= frac{p}{q}
        $$
        as claimed.



        I don't know who showed this first, but this is pretty straightforward.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          You have
          $$
          sum_{n=1}^infty frac{1}{(q+1)^n} = frac{1}{q+1} cdot frac{1}{1-frac{1}{q+1}}= frac{1}{q}
          $$
          using the formula for geometric series applied to $r = frac{1}{q+1}$, recalling that $0< frac{1}{q+1} < 1$.
          Therefore,
          $$
          sum_{n=1}^infty frac{p}{(q+1)^n}= frac{p}{q}
          $$
          as claimed.



          I don't know who showed this first, but this is pretty straightforward.






          share|cite|improve this answer









          $endgroup$



          You have
          $$
          sum_{n=1}^infty frac{1}{(q+1)^n} = frac{1}{q+1} cdot frac{1}{1-frac{1}{q+1}}= frac{1}{q}
          $$
          using the formula for geometric series applied to $r = frac{1}{q+1}$, recalling that $0< frac{1}{q+1} < 1$.
          Therefore,
          $$
          sum_{n=1}^infty frac{p}{(q+1)^n}= frac{p}{q}
          $$
          as claimed.



          I don't know who showed this first, but this is pretty straightforward.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Feb 2 at 6:16









          Clement C.Clement C.

          51.1k34093




          51.1k34093























              1












              $begingroup$

              The formula you wrote is not entirely correct. What you have is
              $$tag1
              frac pq=sum_{k=1}^infty left(frac p{p+q}right)^k.
              $$
              This works nicely when $p,q>0$, since in this case $left|tfrac p{p+q}right|<1$. The formula does not work for many negative numbers.



              As for who may have come up with such formula, the geometric series has been known for millennia: the formula, at least for a finite sum, already appears in Euclid's work. This is basic enough that I doubt it ever appeared as a theorem in anybody's work.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                The formula you wrote is not entirely correct. What you have is
                $$tag1
                frac pq=sum_{k=1}^infty left(frac p{p+q}right)^k.
                $$
                This works nicely when $p,q>0$, since in this case $left|tfrac p{p+q}right|<1$. The formula does not work for many negative numbers.



                As for who may have come up with such formula, the geometric series has been known for millennia: the formula, at least for a finite sum, already appears in Euclid's work. This is basic enough that I doubt it ever appeared as a theorem in anybody's work.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  The formula you wrote is not entirely correct. What you have is
                  $$tag1
                  frac pq=sum_{k=1}^infty left(frac p{p+q}right)^k.
                  $$
                  This works nicely when $p,q>0$, since in this case $left|tfrac p{p+q}right|<1$. The formula does not work for many negative numbers.



                  As for who may have come up with such formula, the geometric series has been known for millennia: the formula, at least for a finite sum, already appears in Euclid's work. This is basic enough that I doubt it ever appeared as a theorem in anybody's work.






                  share|cite|improve this answer









                  $endgroup$



                  The formula you wrote is not entirely correct. What you have is
                  $$tag1
                  frac pq=sum_{k=1}^infty left(frac p{p+q}right)^k.
                  $$
                  This works nicely when $p,q>0$, since in this case $left|tfrac p{p+q}right|<1$. The formula does not work for many negative numbers.



                  As for who may have come up with such formula, the geometric series has been known for millennia: the formula, at least for a finite sum, already appears in Euclid's work. This is basic enough that I doubt it ever appeared as a theorem in anybody's work.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Feb 2 at 14:50









                  Martin ArgeramiMartin Argerami

                  129k1184185




                  129k1184185






























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