Is the sequence of sums of inverse of natural numbers bounded?












4














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? I just started sequences the other day, so forgive me if this is an easy problem.






calculus sequences-and-series harmonic-numbers







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  • 2




    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 $n$?
    – Lord Shark the Unknown
    4 hours ago










  • @LordSharktheUnknown I'm not too sure what that $M$ would be. Can you give a hint?
    – kyle campbell
    3 hours ago






  • 1




    Those aren't subsequences. In any case, are you sure you have transcribed Spivak's sequence correctly?
    – Lord Shark the Unknown
    3 hours ago






  • 1




    It is just divergence of harmonic series. You can search the net for 'divergence of harmonic series'.
    – Kavi Rama Murthy
    3 hours ago






  • 1




    A classic estimate: for each $n$, consider estimating $a_{2^n}$.
    – xbh
    3 hours ago
















4














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? I just started sequences the other day, so forgive me if this is an easy problem.






calculus sequences-and-series harmonic-numbers







share|cite|improve this question




















  • 2




    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 $n$?
    – Lord Shark the Unknown
    4 hours ago










  • @LordSharktheUnknown I'm not too sure what that $M$ would be. Can you give a hint?
    – kyle campbell
    3 hours ago






  • 1




    Those aren't subsequences. In any case, are you sure you have transcribed Spivak's sequence correctly?
    – Lord Shark the Unknown
    3 hours ago






  • 1




    It is just divergence of harmonic series. You can search the net for 'divergence of harmonic series'.
    – Kavi Rama Murthy
    3 hours ago






  • 1




    A classic estimate: for each $n$, consider estimating $a_{2^n}$.
    – xbh
    3 hours ago














4












4








4


1





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? I just started sequences the other day, so forgive me if this is an easy problem.






calculus sequences-and-series harmonic-numbers







share|cite|improve this question















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? I just started sequences the other day, so forgive me if this is an easy problem.






calculus sequences-and-series harmonic-numbers




calculus sequences-and-series harmonic-numbers






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edited 25 mins ago









Arnaud D.

15.7k52443




15.7k52443










asked 4 hours ago









kyle campbellkyle campbell

464




464








  • 2




    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 $n$?
    – Lord Shark the Unknown
    4 hours ago










  • @LordSharktheUnknown I'm not too sure what that $M$ would be. Can you give a hint?
    – kyle campbell
    3 hours ago






  • 1




    Those aren't subsequences. In any case, are you sure you have transcribed Spivak's sequence correctly?
    – Lord Shark the Unknown
    3 hours ago






  • 1




    It is just divergence of harmonic series. You can search the net for 'divergence of harmonic series'.
    – Kavi Rama Murthy
    3 hours ago






  • 1




    A classic estimate: for each $n$, consider estimating $a_{2^n}$.
    – xbh
    3 hours ago














  • 2




    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 $n$?
    – Lord Shark the Unknown
    4 hours ago










  • @LordSharktheUnknown I'm not too sure what that $M$ would be. Can you give a hint?
    – kyle campbell
    3 hours ago






  • 1




    Those aren't subsequences. In any case, are you sure you have transcribed Spivak's sequence correctly?
    – Lord Shark the Unknown
    3 hours ago






  • 1




    It is just divergence of harmonic series. You can search the net for 'divergence of harmonic series'.
    – Kavi Rama Murthy
    3 hours ago






  • 1




    A classic estimate: for each $n$, consider estimating $a_{2^n}$.
    – xbh
    3 hours ago








2




2




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 $n$?
– Lord Shark the Unknown
4 hours ago




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 $n$?
– Lord Shark the Unknown
4 hours ago












@LordSharktheUnknown I'm not too sure what that $M$ would be. Can you give a hint?
– kyle campbell
3 hours ago




@LordSharktheUnknown I'm not too sure what that $M$ would be. Can you give a hint?
– kyle campbell
3 hours ago




1




1




Those aren't subsequences. In any case, are you sure you have transcribed Spivak's sequence correctly?
– Lord Shark the Unknown
3 hours ago




Those aren't subsequences. In any case, are you sure you have transcribed Spivak's sequence correctly?
– Lord Shark the Unknown
3 hours ago




1




1




It is just divergence of harmonic series. You can search the net for 'divergence of harmonic series'.
– Kavi Rama Murthy
3 hours ago




It is just divergence of harmonic series. You can search the net for 'divergence of harmonic series'.
– Kavi Rama Murthy
3 hours ago




1




1




A classic estimate: for each $n$, consider estimating $a_{2^n}$.
– xbh
3 hours ago




A classic estimate: for each $n$, consider estimating $a_{2^n}$.
– xbh
3 hours ago










2 Answers
2






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6














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

















  • 1




    Indeed! Thank you for that input.
    – kyle campbell
    3 hours ago



















4














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.






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    2 Answers
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    2 Answers
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    6














    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

















    • 1




      Indeed! Thank you for that input.
      – kyle campbell
      3 hours ago
















    6














    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

















    • 1




      Indeed! Thank you for that input.
      – kyle campbell
      3 hours ago














    6












    6








    6






    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












    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 3 hours ago









    Tyler6Tyler6

    719212




    719212








    • 1




      Indeed! Thank you for that input.
      – kyle campbell
      3 hours ago














    • 1




      Indeed! Thank you for that input.
      – kyle campbell
      3 hours ago








    1




    1




    Indeed! Thank you for that input.
    – kyle campbell
    3 hours ago




    Indeed! Thank you for that input.
    – kyle campbell
    3 hours ago











    4














    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


























      4














      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
























        4












        4








        4






        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












        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 2 hours ago









        MustangMustang

        3127




        3127






























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