Is the sequence of sums of inverse of natural numbers bounded? [duplicate]












8












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This question already has an answer here:




  • Why does the series $sum_{n=1}^inftyfrac1n$ not converge?

    21 answers




I'm reading through Spivak Ch.22 (Infinite Sequences) right now. He mentioned in the written portion that it's often not a trivial matter to determine the boundedness of sequences. With that in mind, he gave us a sequence to chew on before we learn more about boundedness. That sequence is:



$$1, 1+frac{1}{2}, 1+frac{1}{2}+frac{1}{3}, 1+frac{1}{2}+frac{1}{3}+frac{1}{4}, . . .$$



I know that a sequence is bounded above if there is a number $M$ such that $a_nleq M$ for all $n$. Any hints here?






calculus sequences-and-series harmonic-numbers







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marked as duplicate by Arnaud D., Martin R, user159517, Saad, Matthew Towers Jan 10 at 16:24


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.











  • 3




    $begingroup$
    It is good that you are attending to the definition of boundedness. So, can you write down a suitable $M$ with $a_nle M$ for all $n$?
    $endgroup$
    – Lord Shark the Unknown
    Jan 10 at 5:08










  • $begingroup$
    @LordSharktheUnknown I'm not too sure what that $M$ would be. Can you give a hint?
    $endgroup$
    – kyle campbell
    Jan 10 at 5:14






  • 1




    $begingroup$
    Those aren't subsequences. In any case, are you sure you have transcribed Spivak's sequence correctly?
    $endgroup$
    – Lord Shark the Unknown
    Jan 10 at 5:17






  • 1




    $begingroup$
    It is just divergence of harmonic series. You can search the net for 'divergence of harmonic series'.
    $endgroup$
    – Kavi Rama Murthy
    Jan 10 at 5:23






  • 1




    $begingroup$
    A classic estimate: for each $n$, consider estimating $a_{2^n}$.
    $endgroup$
    – xbh
    Jan 10 at 5:26
















8












$begingroup$



This question already has an answer here:




  • Why does the series $sum_{n=1}^inftyfrac1n$ not converge?

    21 answers




I'm reading through Spivak Ch.22 (Infinite Sequences) right now. He mentioned in the written portion that it's often not a trivial matter to determine the boundedness of sequences. With that in mind, he gave us a sequence to chew on before we learn more about boundedness. That sequence is:



$$1, 1+frac{1}{2}, 1+frac{1}{2}+frac{1}{3}, 1+frac{1}{2}+frac{1}{3}+frac{1}{4}, . . .$$



I know that a sequence is bounded above if there is a number $M$ such that $a_nleq M$ for all $n$. Any hints here?






calculus sequences-and-series harmonic-numbers







share|cite|improve this question











$endgroup$



marked as duplicate by Arnaud D., Martin R, user159517, Saad, Matthew Towers Jan 10 at 16:24


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.











  • 3




    $begingroup$
    It is good that you are attending to the definition of boundedness. So, can you write down a suitable $M$ with $a_nle M$ for all $n$?
    $endgroup$
    – Lord Shark the Unknown
    Jan 10 at 5:08










  • $begingroup$
    @LordSharktheUnknown I'm not too sure what that $M$ would be. Can you give a hint?
    $endgroup$
    – kyle campbell
    Jan 10 at 5:14






  • 1




    $begingroup$
    Those aren't subsequences. In any case, are you sure you have transcribed Spivak's sequence correctly?
    $endgroup$
    – Lord Shark the Unknown
    Jan 10 at 5:17






  • 1




    $begingroup$
    It is just divergence of harmonic series. You can search the net for 'divergence of harmonic series'.
    $endgroup$
    – Kavi Rama Murthy
    Jan 10 at 5:23






  • 1




    $begingroup$
    A classic estimate: for each $n$, consider estimating $a_{2^n}$.
    $endgroup$
    – xbh
    Jan 10 at 5:26














8












8








8


3



$begingroup$



This question already has an answer here:




  • Why does the series $sum_{n=1}^inftyfrac1n$ not converge?

    21 answers




I'm reading through Spivak Ch.22 (Infinite Sequences) right now. He mentioned in the written portion that it's often not a trivial matter to determine the boundedness of sequences. With that in mind, he gave us a sequence to chew on before we learn more about boundedness. That sequence is:



$$1, 1+frac{1}{2}, 1+frac{1}{2}+frac{1}{3}, 1+frac{1}{2}+frac{1}{3}+frac{1}{4}, . . .$$



I know that a sequence is bounded above if there is a number $M$ such that $a_nleq M$ for all $n$. Any hints here?






calculus sequences-and-series harmonic-numbers







share|cite|improve this question











$endgroup$





This question already has an answer here:




  • Why does the series $sum_{n=1}^inftyfrac1n$ not converge?

    21 answers




I'm reading through Spivak Ch.22 (Infinite Sequences) right now. He mentioned in the written portion that it's often not a trivial matter to determine the boundedness of sequences. With that in mind, he gave us a sequence to chew on before we learn more about boundedness. That sequence is:



$$1, 1+frac{1}{2}, 1+frac{1}{2}+frac{1}{3}, 1+frac{1}{2}+frac{1}{3}+frac{1}{4}, . . .$$



I know that a sequence is bounded above if there is a number $M$ such that $a_nleq M$ for all $n$. Any hints here?





This question already has an answer here:




  • Why does the series $sum_{n=1}^inftyfrac1n$ not converge?

    21 answers







calculus sequences-and-series harmonic-numbers




calculus sequences-and-series harmonic-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 11 at 4:02







kyle campbell

















asked Jan 10 at 5:03









kyle campbellkyle campbell

725




725




marked as duplicate by Arnaud D., Martin R, user159517, Saad, Matthew Towers Jan 10 at 16:24


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by Arnaud D., Martin R, user159517, Saad, Matthew Towers Jan 10 at 16:24


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 3




    $begingroup$
    It is good that you are attending to the definition of boundedness. So, can you write down a suitable $M$ with $a_nle M$ for all $n$?
    $endgroup$
    – Lord Shark the Unknown
    Jan 10 at 5:08










  • $begingroup$
    @LordSharktheUnknown I'm not too sure what that $M$ would be. Can you give a hint?
    $endgroup$
    – kyle campbell
    Jan 10 at 5:14






  • 1




    $begingroup$
    Those aren't subsequences. In any case, are you sure you have transcribed Spivak's sequence correctly?
    $endgroup$
    – Lord Shark the Unknown
    Jan 10 at 5:17






  • 1




    $begingroup$
    It is just divergence of harmonic series. You can search the net for 'divergence of harmonic series'.
    $endgroup$
    – Kavi Rama Murthy
    Jan 10 at 5:23






  • 1




    $begingroup$
    A classic estimate: for each $n$, consider estimating $a_{2^n}$.
    $endgroup$
    – xbh
    Jan 10 at 5:26














  • 3




    $begingroup$
    It is good that you are attending to the definition of boundedness. So, can you write down a suitable $M$ with $a_nle M$ for all $n$?
    $endgroup$
    – Lord Shark the Unknown
    Jan 10 at 5:08










  • $begingroup$
    @LordSharktheUnknown I'm not too sure what that $M$ would be. Can you give a hint?
    $endgroup$
    – kyle campbell
    Jan 10 at 5:14






  • 1




    $begingroup$
    Those aren't subsequences. In any case, are you sure you have transcribed Spivak's sequence correctly?
    $endgroup$
    – Lord Shark the Unknown
    Jan 10 at 5:17






  • 1




    $begingroup$
    It is just divergence of harmonic series. You can search the net for 'divergence of harmonic series'.
    $endgroup$
    – Kavi Rama Murthy
    Jan 10 at 5:23






  • 1




    $begingroup$
    A classic estimate: for each $n$, consider estimating $a_{2^n}$.
    $endgroup$
    – xbh
    Jan 10 at 5:26








3




3




$begingroup$
It is good that you are attending to the definition of boundedness. So, can you write down a suitable $M$ with $a_nle M$ for all $n$?
$endgroup$
– Lord Shark the Unknown
Jan 10 at 5:08




$begingroup$
It is good that you are attending to the definition of boundedness. So, can you write down a suitable $M$ with $a_nle M$ for all $n$?
$endgroup$
– Lord Shark the Unknown
Jan 10 at 5:08












$begingroup$
@LordSharktheUnknown I'm not too sure what that $M$ would be. Can you give a hint?
$endgroup$
– kyle campbell
Jan 10 at 5:14




$begingroup$
@LordSharktheUnknown I'm not too sure what that $M$ would be. Can you give a hint?
$endgroup$
– kyle campbell
Jan 10 at 5:14




1




1




$begingroup$
Those aren't subsequences. In any case, are you sure you have transcribed Spivak's sequence correctly?
$endgroup$
– Lord Shark the Unknown
Jan 10 at 5:17




$begingroup$
Those aren't subsequences. In any case, are you sure you have transcribed Spivak's sequence correctly?
$endgroup$
– Lord Shark the Unknown
Jan 10 at 5:17




1




1




$begingroup$
It is just divergence of harmonic series. You can search the net for 'divergence of harmonic series'.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 5:23




$begingroup$
It is just divergence of harmonic series. You can search the net for 'divergence of harmonic series'.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 5:23




1




1




$begingroup$
A classic estimate: for each $n$, consider estimating $a_{2^n}$.
$endgroup$
– xbh
Jan 10 at 5:26




$begingroup$
A classic estimate: for each $n$, consider estimating $a_{2^n}$.
$endgroup$
– xbh
Jan 10 at 5:26










2 Answers
2






active

oldest

votes


















14












$begingroup$

This is a classic example of how our intuition can be wrong concerning infinities. As @Kavi said in the comments, this is the harmonic series and indeed diverges. I believe that Spivak provided this as an example to “chew on” in order to show how non trivial boundedness can be, as on first glance many people believe this sequence should be bounded.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Indeed! Thank you for that input.
    $endgroup$
    – kyle campbell
    Jan 10 at 5:57






  • 3




    $begingroup$
    This is not the harmonic series, it is the sequence of partial sums of the harmonic series. A sequence and a series are two different things. Other important sequences linked to a series are the sequence of the terms (${1/n}$ in this case) and the sequence of partial rests.
    $endgroup$
    – A. Arredondo
    Jan 10 at 9:16



















8












$begingroup$

Indeed, you can notice that



$$a_1=1,a_4>2(=1+frac{1}{2}+frac{1}{4}+frac{1}{4}), a_8>frac{5}{2}(=a_4+frac{1}{8}+frac{1}{8}+frac{1}{8}+frac{1}{8}),a_{16}>3(=a_{8}+frac{1}{16}+frac{1}{16}+..8times)$$, so the general pattern is $a_{2^{2n}}>n+1$, so given an upper bound M, you can find a natural number n s.t. $ngeq M$ by archimedian property and so $a_{2^{2n}}>n+1>M$ and hence it can't be bounded.






share|cite|improve this answer









$endgroup$




















    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    14












    $begingroup$

    This is a classic example of how our intuition can be wrong concerning infinities. As @Kavi said in the comments, this is the harmonic series and indeed diverges. I believe that Spivak provided this as an example to “chew on” in order to show how non trivial boundedness can be, as on first glance many people believe this sequence should be bounded.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      Indeed! Thank you for that input.
      $endgroup$
      – kyle campbell
      Jan 10 at 5:57






    • 3




      $begingroup$
      This is not the harmonic series, it is the sequence of partial sums of the harmonic series. A sequence and a series are two different things. Other important sequences linked to a series are the sequence of the terms (${1/n}$ in this case) and the sequence of partial rests.
      $endgroup$
      – A. Arredondo
      Jan 10 at 9:16
















    14












    $begingroup$

    This is a classic example of how our intuition can be wrong concerning infinities. As @Kavi said in the comments, this is the harmonic series and indeed diverges. I believe that Spivak provided this as an example to “chew on” in order to show how non trivial boundedness can be, as on first glance many people believe this sequence should be bounded.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      Indeed! Thank you for that input.
      $endgroup$
      – kyle campbell
      Jan 10 at 5:57






    • 3




      $begingroup$
      This is not the harmonic series, it is the sequence of partial sums of the harmonic series. A sequence and a series are two different things. Other important sequences linked to a series are the sequence of the terms (${1/n}$ in this case) and the sequence of partial rests.
      $endgroup$
      – A. Arredondo
      Jan 10 at 9:16














    14












    14








    14





    $begingroup$

    This is a classic example of how our intuition can be wrong concerning infinities. As @Kavi said in the comments, this is the harmonic series and indeed diverges. I believe that Spivak provided this as an example to “chew on” in order to show how non trivial boundedness can be, as on first glance many people believe this sequence should be bounded.






    share|cite|improve this answer









    $endgroup$



    This is a classic example of how our intuition can be wrong concerning infinities. As @Kavi said in the comments, this is the harmonic series and indeed diverges. I believe that Spivak provided this as an example to “chew on” in order to show how non trivial boundedness can be, as on first glance many people believe this sequence should be bounded.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 10 at 5:48









    Tyler6Tyler6

    836313




    836313








    • 1




      $begingroup$
      Indeed! Thank you for that input.
      $endgroup$
      – kyle campbell
      Jan 10 at 5:57






    • 3




      $begingroup$
      This is not the harmonic series, it is the sequence of partial sums of the harmonic series. A sequence and a series are two different things. Other important sequences linked to a series are the sequence of the terms (${1/n}$ in this case) and the sequence of partial rests.
      $endgroup$
      – A. Arredondo
      Jan 10 at 9:16














    • 1




      $begingroup$
      Indeed! Thank you for that input.
      $endgroup$
      – kyle campbell
      Jan 10 at 5:57






    • 3




      $begingroup$
      This is not the harmonic series, it is the sequence of partial sums of the harmonic series. A sequence and a series are two different things. Other important sequences linked to a series are the sequence of the terms (${1/n}$ in this case) and the sequence of partial rests.
      $endgroup$
      – A. Arredondo
      Jan 10 at 9:16








    1




    1




    $begingroup$
    Indeed! Thank you for that input.
    $endgroup$
    – kyle campbell
    Jan 10 at 5:57




    $begingroup$
    Indeed! Thank you for that input.
    $endgroup$
    – kyle campbell
    Jan 10 at 5:57




    3




    3




    $begingroup$
    This is not the harmonic series, it is the sequence of partial sums of the harmonic series. A sequence and a series are two different things. Other important sequences linked to a series are the sequence of the terms (${1/n}$ in this case) and the sequence of partial rests.
    $endgroup$
    – A. Arredondo
    Jan 10 at 9:16




    $begingroup$
    This is not the harmonic series, it is the sequence of partial sums of the harmonic series. A sequence and a series are two different things. Other important sequences linked to a series are the sequence of the terms (${1/n}$ in this case) and the sequence of partial rests.
    $endgroup$
    – A. Arredondo
    Jan 10 at 9:16











    8












    $begingroup$

    Indeed, you can notice that



    $$a_1=1,a_4>2(=1+frac{1}{2}+frac{1}{4}+frac{1}{4}), a_8>frac{5}{2}(=a_4+frac{1}{8}+frac{1}{8}+frac{1}{8}+frac{1}{8}),a_{16}>3(=a_{8}+frac{1}{16}+frac{1}{16}+..8times)$$, so the general pattern is $a_{2^{2n}}>n+1$, so given an upper bound M, you can find a natural number n s.t. $ngeq M$ by archimedian property and so $a_{2^{2n}}>n+1>M$ and hence it can't be bounded.






    share|cite|improve this answer









    $endgroup$


















      8












      $begingroup$

      Indeed, you can notice that



      $$a_1=1,a_4>2(=1+frac{1}{2}+frac{1}{4}+frac{1}{4}), a_8>frac{5}{2}(=a_4+frac{1}{8}+frac{1}{8}+frac{1}{8}+frac{1}{8}),a_{16}>3(=a_{8}+frac{1}{16}+frac{1}{16}+..8times)$$, so the general pattern is $a_{2^{2n}}>n+1$, so given an upper bound M, you can find a natural number n s.t. $ngeq M$ by archimedian property and so $a_{2^{2n}}>n+1>M$ and hence it can't be bounded.






      share|cite|improve this answer









      $endgroup$
















        8












        8








        8





        $begingroup$

        Indeed, you can notice that



        $$a_1=1,a_4>2(=1+frac{1}{2}+frac{1}{4}+frac{1}{4}), a_8>frac{5}{2}(=a_4+frac{1}{8}+frac{1}{8}+frac{1}{8}+frac{1}{8}),a_{16}>3(=a_{8}+frac{1}{16}+frac{1}{16}+..8times)$$, so the general pattern is $a_{2^{2n}}>n+1$, so given an upper bound M, you can find a natural number n s.t. $ngeq M$ by archimedian property and so $a_{2^{2n}}>n+1>M$ and hence it can't be bounded.






        share|cite|improve this answer









        $endgroup$



        Indeed, you can notice that



        $$a_1=1,a_4>2(=1+frac{1}{2}+frac{1}{4}+frac{1}{4}), a_8>frac{5}{2}(=a_4+frac{1}{8}+frac{1}{8}+frac{1}{8}+frac{1}{8}),a_{16}>3(=a_{8}+frac{1}{16}+frac{1}{16}+..8times)$$, so the general pattern is $a_{2^{2n}}>n+1$, so given an upper bound M, you can find a natural number n s.t. $ngeq M$ by archimedian property and so $a_{2^{2n}}>n+1>M$ and hence it can't be bounded.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 10 at 6:34









        MustangMustang

        3407




        3407















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