Two sums, the relation between them












0














What is the relation between sum? Is something interesting? Is it possible taking square or solving set of equation calculate "non squared" sum? Thank you for your explanation.



$$sumlimits_{n=2}^{infty}(frac{1}{n}-frac{1}{n-1})=-1$$



and



$$sumlimits_{n=2}^{infty}(frac{1}{n^2}-frac{1}{(n-1)^2})=frac{pi^2}{6}-1$$






real-analysis integration sequences-and-series number-theory power-series







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  • The second sum looks wrong. For one thing, the value should be negative...
    – Erick Wong
    Oct 19 '18 at 2:42
















0














What is the relation between sum? Is something interesting? Is it possible taking square or solving set of equation calculate "non squared" sum? Thank you for your explanation.



$$sumlimits_{n=2}^{infty}(frac{1}{n}-frac{1}{n-1})=-1$$



and



$$sumlimits_{n=2}^{infty}(frac{1}{n^2}-frac{1}{(n-1)^2})=frac{pi^2}{6}-1$$






real-analysis integration sequences-and-series number-theory power-series







share|cite|improve this question
























  • The second sum looks wrong. For one thing, the value should be negative...
    – Erick Wong
    Oct 19 '18 at 2:42














0












0








0


0





What is the relation between sum? Is something interesting? Is it possible taking square or solving set of equation calculate "non squared" sum? Thank you for your explanation.



$$sumlimits_{n=2}^{infty}(frac{1}{n}-frac{1}{n-1})=-1$$



and



$$sumlimits_{n=2}^{infty}(frac{1}{n^2}-frac{1}{(n-1)^2})=frac{pi^2}{6}-1$$






real-analysis integration sequences-and-series number-theory power-series







share|cite|improve this question















What is the relation between sum? Is something interesting? Is it possible taking square or solving set of equation calculate "non squared" sum? Thank you for your explanation.



$$sumlimits_{n=2}^{infty}(frac{1}{n}-frac{1}{n-1})=-1$$



and



$$sumlimits_{n=2}^{infty}(frac{1}{n^2}-frac{1}{(n-1)^2})=frac{pi^2}{6}-1$$






real-analysis integration sequences-and-series number-theory power-series




real-analysis integration sequences-and-series number-theory power-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Oct 19 '18 at 12:20







Marianna Kalwat

















asked Oct 19 '18 at 1:55









Marianna KalwatMarianna Kalwat

626




626












  • The second sum looks wrong. For one thing, the value should be negative...
    – Erick Wong
    Oct 19 '18 at 2:42


















  • The second sum looks wrong. For one thing, the value should be negative...
    – Erick Wong
    Oct 19 '18 at 2:42
















The second sum looks wrong. For one thing, the value should be negative...
– Erick Wong
Oct 19 '18 at 2:42




The second sum looks wrong. For one thing, the value should be negative...
– Erick Wong
Oct 19 '18 at 2:42










2 Answers
2






active

oldest

votes


















0














I do not understand much what you mean but maybe these points help you in something



1)The last result is wrong, first think that you can separate this sum into two parts
$$sumlimits_{n=2}^{infty}(frac{1}{n^2}-frac{1}{(n-1)^2})=sumlimits_{n=2}^{infty}frac{1}{n^2}-sumlimits_{n=2}^{infty}{frac{1}{(n-1)^2}}$$



then you can see that
$$sumlimits_{n=2}^{infty}frac{1}{n^2}=sumlimits_{n=1}^{infty}frac{1}{n^2}-(frac{1}{(n=1)^2})=sumlimits_{n=1}^{infty}frac{1}{n^2}-1$$



and at the same time that



$$sumlimits_{n=2}^{infty}{frac{1}{(n-1)^2}}=sumlimits_{n=1}^{infty}frac{1}{n^2}$$



then using what has already been said and this sum ,discovered by euler,$
sumlimits_{n=1}^{infty}frac{1}{n^2}=frac{pi^2}{6}$ We can replace it and gives us



$$sumlimits_{n=2}^{infty}frac{1}{n^2}-sumlimits_{n=2}^{infty}{frac{1}{(n-1)^2}}=frac{pi^2}{6}-1-frac{pi^2}{6}=-1$$



2)Also in the same way we can generalize these sums for any exponent



$$sumlimits_{n=2}^{infty}(frac{1}{n^k}-frac{1}{(n-1)^k})$$



Applying the same steps as before we get



$$sumlimits_{n=2}^{infty}(frac{1}{n^k}-frac{1}{(n-1)^k})=sumlimits_{n=2}^{infty}frac{1}{n^k}-sumlimits_{n=2}^{infty}{frac{1}{(n-1)^k}}=zeta(k)-1-zeta(k)=-1$$



Thanks you for reading.






share|cite|improve this answer





















  • How you get $ zeta(k)-1-zeta(k)$?
    – Marianna Kalwat
    Oct 19 '18 at 12:26












  • we have to do the same steps, keep in mind that $sumlimits_{n=2}^{infty}frac{1}{n^k}=sumlimits_{n=1}^{infty}frac{1}{n^k}-frac{1}{({n^k=1})}=zeta(k)-1$ and at the same time $sumlimits_{n=2}^{infty}frac{1}{(n-1)^k}=zeta(k)$
    – Blas Casazza
    Oct 19 '18 at 17:13





















0














In fact we have $$sum_{n=1}^{infty}a_{n+1}-a_n=(a_2-a_1)+(a_3-a_2)+(a_4-a_3)+cdots=a_{infty}-a_1$$
These interesting sequences are referred to as Telescoping series. Also the second series has been calculated wrong.



$$sumlimits_{n=2}^{infty}(frac{1}{n^2}-frac{1}{(n-1)^2})=-1$$






share|cite|improve this answer





















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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

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    active

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    0














    I do not understand much what you mean but maybe these points help you in something



    1)The last result is wrong, first think that you can separate this sum into two parts
    $$sumlimits_{n=2}^{infty}(frac{1}{n^2}-frac{1}{(n-1)^2})=sumlimits_{n=2}^{infty}frac{1}{n^2}-sumlimits_{n=2}^{infty}{frac{1}{(n-1)^2}}$$



    then you can see that
    $$sumlimits_{n=2}^{infty}frac{1}{n^2}=sumlimits_{n=1}^{infty}frac{1}{n^2}-(frac{1}{(n=1)^2})=sumlimits_{n=1}^{infty}frac{1}{n^2}-1$$



    and at the same time that



    $$sumlimits_{n=2}^{infty}{frac{1}{(n-1)^2}}=sumlimits_{n=1}^{infty}frac{1}{n^2}$$



    then using what has already been said and this sum ,discovered by euler,$
    sumlimits_{n=1}^{infty}frac{1}{n^2}=frac{pi^2}{6}$     We can replace it and gives us



    $$sumlimits_{n=2}^{infty}frac{1}{n^2}-sumlimits_{n=2}^{infty}{frac{1}{(n-1)^2}}=frac{pi^2}{6}-1-frac{pi^2}{6}=-1$$



    2)Also in the same way we can generalize these sums for any exponent



    $$sumlimits_{n=2}^{infty}(frac{1}{n^k}-frac{1}{(n-1)^k})$$



    Applying the same steps as before we get



    $$sumlimits_{n=2}^{infty}(frac{1}{n^k}-frac{1}{(n-1)^k})=sumlimits_{n=2}^{infty}frac{1}{n^k}-sumlimits_{n=2}^{infty}{frac{1}{(n-1)^k}}=zeta(k)-1-zeta(k)=-1$$



    Thanks you for reading.






    share|cite|improve this answer





















    • How you get $ zeta(k)-1-zeta(k)$?
      – Marianna Kalwat
      Oct 19 '18 at 12:26












    • we have to do the same steps, keep in mind that $sumlimits_{n=2}^{infty}frac{1}{n^k}=sumlimits_{n=1}^{infty}frac{1}{n^k}-frac{1}{({n^k=1})}=zeta(k)-1$ and at the same time $sumlimits_{n=2}^{infty}frac{1}{(n-1)^k}=zeta(k)$
      – Blas Casazza
      Oct 19 '18 at 17:13


















    0














    I do not understand much what you mean but maybe these points help you in something



    1)The last result is wrong, first think that you can separate this sum into two parts
    $$sumlimits_{n=2}^{infty}(frac{1}{n^2}-frac{1}{(n-1)^2})=sumlimits_{n=2}^{infty}frac{1}{n^2}-sumlimits_{n=2}^{infty}{frac{1}{(n-1)^2}}$$



    then you can see that
    $$sumlimits_{n=2}^{infty}frac{1}{n^2}=sumlimits_{n=1}^{infty}frac{1}{n^2}-(frac{1}{(n=1)^2})=sumlimits_{n=1}^{infty}frac{1}{n^2}-1$$



    and at the same time that



    $$sumlimits_{n=2}^{infty}{frac{1}{(n-1)^2}}=sumlimits_{n=1}^{infty}frac{1}{n^2}$$



    then using what has already been said and this sum ,discovered by euler,$
    sumlimits_{n=1}^{infty}frac{1}{n^2}=frac{pi^2}{6}$     We can replace it and gives us



    $$sumlimits_{n=2}^{infty}frac{1}{n^2}-sumlimits_{n=2}^{infty}{frac{1}{(n-1)^2}}=frac{pi^2}{6}-1-frac{pi^2}{6}=-1$$



    2)Also in the same way we can generalize these sums for any exponent



    $$sumlimits_{n=2}^{infty}(frac{1}{n^k}-frac{1}{(n-1)^k})$$



    Applying the same steps as before we get



    $$sumlimits_{n=2}^{infty}(frac{1}{n^k}-frac{1}{(n-1)^k})=sumlimits_{n=2}^{infty}frac{1}{n^k}-sumlimits_{n=2}^{infty}{frac{1}{(n-1)^k}}=zeta(k)-1-zeta(k)=-1$$



    Thanks you for reading.






    share|cite|improve this answer





















    • How you get $ zeta(k)-1-zeta(k)$?
      – Marianna Kalwat
      Oct 19 '18 at 12:26












    • we have to do the same steps, keep in mind that $sumlimits_{n=2}^{infty}frac{1}{n^k}=sumlimits_{n=1}^{infty}frac{1}{n^k}-frac{1}{({n^k=1})}=zeta(k)-1$ and at the same time $sumlimits_{n=2}^{infty}frac{1}{(n-1)^k}=zeta(k)$
      – Blas Casazza
      Oct 19 '18 at 17:13
















    0












    0








    0






    I do not understand much what you mean but maybe these points help you in something



    1)The last result is wrong, first think that you can separate this sum into two parts
    $$sumlimits_{n=2}^{infty}(frac{1}{n^2}-frac{1}{(n-1)^2})=sumlimits_{n=2}^{infty}frac{1}{n^2}-sumlimits_{n=2}^{infty}{frac{1}{(n-1)^2}}$$



    then you can see that
    $$sumlimits_{n=2}^{infty}frac{1}{n^2}=sumlimits_{n=1}^{infty}frac{1}{n^2}-(frac{1}{(n=1)^2})=sumlimits_{n=1}^{infty}frac{1}{n^2}-1$$



    and at the same time that



    $$sumlimits_{n=2}^{infty}{frac{1}{(n-1)^2}}=sumlimits_{n=1}^{infty}frac{1}{n^2}$$



    then using what has already been said and this sum ,discovered by euler,$
    sumlimits_{n=1}^{infty}frac{1}{n^2}=frac{pi^2}{6}$     We can replace it and gives us



    $$sumlimits_{n=2}^{infty}frac{1}{n^2}-sumlimits_{n=2}^{infty}{frac{1}{(n-1)^2}}=frac{pi^2}{6}-1-frac{pi^2}{6}=-1$$



    2)Also in the same way we can generalize these sums for any exponent



    $$sumlimits_{n=2}^{infty}(frac{1}{n^k}-frac{1}{(n-1)^k})$$



    Applying the same steps as before we get



    $$sumlimits_{n=2}^{infty}(frac{1}{n^k}-frac{1}{(n-1)^k})=sumlimits_{n=2}^{infty}frac{1}{n^k}-sumlimits_{n=2}^{infty}{frac{1}{(n-1)^k}}=zeta(k)-1-zeta(k)=-1$$



    Thanks you for reading.






    share|cite|improve this answer












    I do not understand much what you mean but maybe these points help you in something



    1)The last result is wrong, first think that you can separate this sum into two parts
    $$sumlimits_{n=2}^{infty}(frac{1}{n^2}-frac{1}{(n-1)^2})=sumlimits_{n=2}^{infty}frac{1}{n^2}-sumlimits_{n=2}^{infty}{frac{1}{(n-1)^2}}$$



    then you can see that
    $$sumlimits_{n=2}^{infty}frac{1}{n^2}=sumlimits_{n=1}^{infty}frac{1}{n^2}-(frac{1}{(n=1)^2})=sumlimits_{n=1}^{infty}frac{1}{n^2}-1$$



    and at the same time that



    $$sumlimits_{n=2}^{infty}{frac{1}{(n-1)^2}}=sumlimits_{n=1}^{infty}frac{1}{n^2}$$



    then using what has already been said and this sum ,discovered by euler,$
    sumlimits_{n=1}^{infty}frac{1}{n^2}=frac{pi^2}{6}$     We can replace it and gives us



    $$sumlimits_{n=2}^{infty}frac{1}{n^2}-sumlimits_{n=2}^{infty}{frac{1}{(n-1)^2}}=frac{pi^2}{6}-1-frac{pi^2}{6}=-1$$



    2)Also in the same way we can generalize these sums for any exponent



    $$sumlimits_{n=2}^{infty}(frac{1}{n^k}-frac{1}{(n-1)^k})$$



    Applying the same steps as before we get



    $$sumlimits_{n=2}^{infty}(frac{1}{n^k}-frac{1}{(n-1)^k})=sumlimits_{n=2}^{infty}frac{1}{n^k}-sumlimits_{n=2}^{infty}{frac{1}{(n-1)^k}}=zeta(k)-1-zeta(k)=-1$$



    Thanks you for reading.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Oct 19 '18 at 6:20









    Blas CasazzaBlas Casazza

    527




    527












    • How you get $ zeta(k)-1-zeta(k)$?
      – Marianna Kalwat
      Oct 19 '18 at 12:26












    • we have to do the same steps, keep in mind that $sumlimits_{n=2}^{infty}frac{1}{n^k}=sumlimits_{n=1}^{infty}frac{1}{n^k}-frac{1}{({n^k=1})}=zeta(k)-1$ and at the same time $sumlimits_{n=2}^{infty}frac{1}{(n-1)^k}=zeta(k)$
      – Blas Casazza
      Oct 19 '18 at 17:13




















    • How you get $ zeta(k)-1-zeta(k)$?
      – Marianna Kalwat
      Oct 19 '18 at 12:26












    • we have to do the same steps, keep in mind that $sumlimits_{n=2}^{infty}frac{1}{n^k}=sumlimits_{n=1}^{infty}frac{1}{n^k}-frac{1}{({n^k=1})}=zeta(k)-1$ and at the same time $sumlimits_{n=2}^{infty}frac{1}{(n-1)^k}=zeta(k)$
      – Blas Casazza
      Oct 19 '18 at 17:13


















    How you get $ zeta(k)-1-zeta(k)$?
    – Marianna Kalwat
    Oct 19 '18 at 12:26






    How you get $ zeta(k)-1-zeta(k)$?
    – Marianna Kalwat
    Oct 19 '18 at 12:26














    we have to do the same steps, keep in mind that $sumlimits_{n=2}^{infty}frac{1}{n^k}=sumlimits_{n=1}^{infty}frac{1}{n^k}-frac{1}{({n^k=1})}=zeta(k)-1$ and at the same time $sumlimits_{n=2}^{infty}frac{1}{(n-1)^k}=zeta(k)$
    – Blas Casazza
    Oct 19 '18 at 17:13






    we have to do the same steps, keep in mind that $sumlimits_{n=2}^{infty}frac{1}{n^k}=sumlimits_{n=1}^{infty}frac{1}{n^k}-frac{1}{({n^k=1})}=zeta(k)-1$ and at the same time $sumlimits_{n=2}^{infty}frac{1}{(n-1)^k}=zeta(k)$
    – Blas Casazza
    Oct 19 '18 at 17:13













    0














    In fact we have $$sum_{n=1}^{infty}a_{n+1}-a_n=(a_2-a_1)+(a_3-a_2)+(a_4-a_3)+cdots=a_{infty}-a_1$$
    These interesting sequences are referred to as Telescoping series. Also the second series has been calculated wrong.



    $$sumlimits_{n=2}^{infty}(frac{1}{n^2}-frac{1}{(n-1)^2})=-1$$






    share|cite|improve this answer


























      0














      In fact we have $$sum_{n=1}^{infty}a_{n+1}-a_n=(a_2-a_1)+(a_3-a_2)+(a_4-a_3)+cdots=a_{infty}-a_1$$
      These interesting sequences are referred to as Telescoping series. Also the second series has been calculated wrong.



      $$sumlimits_{n=2}^{infty}(frac{1}{n^2}-frac{1}{(n-1)^2})=-1$$






      share|cite|improve this answer
























        0












        0








        0






        In fact we have $$sum_{n=1}^{infty}a_{n+1}-a_n=(a_2-a_1)+(a_3-a_2)+(a_4-a_3)+cdots=a_{infty}-a_1$$
        These interesting sequences are referred to as Telescoping series. Also the second series has been calculated wrong.



        $$sumlimits_{n=2}^{infty}(frac{1}{n^2}-frac{1}{(n-1)^2})=-1$$






        share|cite|improve this answer












        In fact we have $$sum_{n=1}^{infty}a_{n+1}-a_n=(a_2-a_1)+(a_3-a_2)+(a_4-a_3)+cdots=a_{infty}-a_1$$
        These interesting sequences are referred to as Telescoping series. Also the second series has been calculated wrong.



        $$sumlimits_{n=2}^{infty}(frac{1}{n^2}-frac{1}{(n-1)^2})=-1$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 30 '18 at 19:20









        Mostafa AyazMostafa Ayaz

        14.5k3937




        14.5k3937






























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