How can we formally state a series converges?












1














If the series $sum_{n=0}^{infty}na_{n}$ converges absolutely, how can we write this using the definition of convergence?



The definition of convergence states this:



"The series $sum_{n=1}^{infty}a_{n}$ converges if and only if, given $epsilon gt 0$, there exists an $N in mathbf N$ such that whenever $n gt m ge N $ it follows that: |$ a_{m+1} + a_{m + 2} + ... + a_{n}| lt epsilon$ "



How do I extend this to my current series above? What is tricking me and I'm not too sure if what I'm doing is correct is the $n$ infront of the $a_{n}$. If that series converges absolutely, does it imply this:



|$ na_{m+1} + na_{m + 2} + ... + na_{n}| lt epsilon$ ?



Any help would be appreciated.






real-analysis sequences-and-series convergence







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  • Converges absolutely is a stronger condition than converges. What you have written implies convergence but not absolute convergence.
    – Mason
    Nov 18 at 23:08








  • 1




    What you quote is the condition for the sequence of partial sums being Cauchy. In the real numbers this happens to be the same sequences as the convergent ones, but I think it is rare to declare it to be the definition of "converge".
    – Henning Makholm
    Nov 18 at 23:14


















1














If the series $sum_{n=0}^{infty}na_{n}$ converges absolutely, how can we write this using the definition of convergence?



The definition of convergence states this:



"The series $sum_{n=1}^{infty}a_{n}$ converges if and only if, given $epsilon gt 0$, there exists an $N in mathbf N$ such that whenever $n gt m ge N $ it follows that: |$ a_{m+1} + a_{m + 2} + ... + a_{n}| lt epsilon$ "



How do I extend this to my current series above? What is tricking me and I'm not too sure if what I'm doing is correct is the $n$ infront of the $a_{n}$. If that series converges absolutely, does it imply this:



|$ na_{m+1} + na_{m + 2} + ... + na_{n}| lt epsilon$ ?



Any help would be appreciated.






real-analysis sequences-and-series convergence







share|cite|improve this question






















  • Converges absolutely is a stronger condition than converges. What you have written implies convergence but not absolute convergence.
    – Mason
    Nov 18 at 23:08








  • 1




    What you quote is the condition for the sequence of partial sums being Cauchy. In the real numbers this happens to be the same sequences as the convergent ones, but I think it is rare to declare it to be the definition of "converge".
    – Henning Makholm
    Nov 18 at 23:14
















1












1








1







If the series $sum_{n=0}^{infty}na_{n}$ converges absolutely, how can we write this using the definition of convergence?



The definition of convergence states this:



"The series $sum_{n=1}^{infty}a_{n}$ converges if and only if, given $epsilon gt 0$, there exists an $N in mathbf N$ such that whenever $n gt m ge N $ it follows that: |$ a_{m+1} + a_{m + 2} + ... + a_{n}| lt epsilon$ "



How do I extend this to my current series above? What is tricking me and I'm not too sure if what I'm doing is correct is the $n$ infront of the $a_{n}$. If that series converges absolutely, does it imply this:



|$ na_{m+1} + na_{m + 2} + ... + na_{n}| lt epsilon$ ?



Any help would be appreciated.






real-analysis sequences-and-series convergence







share|cite|improve this question













If the series $sum_{n=0}^{infty}na_{n}$ converges absolutely, how can we write this using the definition of convergence?



The definition of convergence states this:



"The series $sum_{n=1}^{infty}a_{n}$ converges if and only if, given $epsilon gt 0$, there exists an $N in mathbf N$ such that whenever $n gt m ge N $ it follows that: |$ a_{m+1} + a_{m + 2} + ... + a_{n}| lt epsilon$ "



How do I extend this to my current series above? What is tricking me and I'm not too sure if what I'm doing is correct is the $n$ infront of the $a_{n}$. If that series converges absolutely, does it imply this:



|$ na_{m+1} + na_{m + 2} + ... + na_{n}| lt epsilon$ ?



Any help would be appreciated.






real-analysis sequences-and-series convergence




real-analysis sequences-and-series convergence






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asked Nov 18 at 23:02









Wallace

1226




1226












  • Converges absolutely is a stronger condition than converges. What you have written implies convergence but not absolute convergence.
    – Mason
    Nov 18 at 23:08








  • 1




    What you quote is the condition for the sequence of partial sums being Cauchy. In the real numbers this happens to be the same sequences as the convergent ones, but I think it is rare to declare it to be the definition of "converge".
    – Henning Makholm
    Nov 18 at 23:14




















  • Converges absolutely is a stronger condition than converges. What you have written implies convergence but not absolute convergence.
    – Mason
    Nov 18 at 23:08








  • 1




    What you quote is the condition for the sequence of partial sums being Cauchy. In the real numbers this happens to be the same sequences as the convergent ones, but I think it is rare to declare it to be the definition of "converge".
    – Henning Makholm
    Nov 18 at 23:14


















Converges absolutely is a stronger condition than converges. What you have written implies convergence but not absolute convergence.
– Mason
Nov 18 at 23:08






Converges absolutely is a stronger condition than converges. What you have written implies convergence but not absolute convergence.
– Mason
Nov 18 at 23:08






1




1




What you quote is the condition for the sequence of partial sums being Cauchy. In the real numbers this happens to be the same sequences as the convergent ones, but I think it is rare to declare it to be the definition of "converge".
– Henning Makholm
Nov 18 at 23:14






What you quote is the condition for the sequence of partial sums being Cauchy. In the real numbers this happens to be the same sequences as the convergent ones, but I think it is rare to declare it to be the definition of "converge".
– Henning Makholm
Nov 18 at 23:14












2 Answers
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The correct inequality is $(m+1)|a_{m+1}|+(m+2)|a_{m+2}|+cdots+n|a_n|<epsilon$.






share|cite|improve this answer































    2














    The series $sum_{k=0}^infty b_k$ converges absolutely $iff$ The series $sum_{k=0}^infty |b_k|$ converges.



    In your case, take $b_k = ka_k$ and plug it into whatever your definition is for (series) convergence.



    Usually, the definition of (series) convergence is the following: The series $sum_{k=0}^infty b_k$ converges to a number $L$ $iff$ "For all $epsilon > 0$, there exists an $N > 0$ such that $n geq N$ implies $left| sum_{k=1}^n b_k - L right| < epsilon$."



    As Henning Makholm correctly said in the comments, the definition you gave is for a series being Cauchy. In the case where the terms of your series are all real numbers, a series being Cauchy is equivalent to a series being convergent.






    share|cite|improve this answer























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      2 Answers
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      2 Answers
      2






      active

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      active

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      0














      The correct inequality is $(m+1)|a_{m+1}|+(m+2)|a_{m+2}|+cdots+n|a_n|<epsilon$.






      share|cite|improve this answer




























        0














        The correct inequality is $(m+1)|a_{m+1}|+(m+2)|a_{m+2}|+cdots+n|a_n|<epsilon$.






        share|cite|improve this answer


























          0












          0








          0






          The correct inequality is $(m+1)|a_{m+1}|+(m+2)|a_{m+2}|+cdots+n|a_n|<epsilon$.






          share|cite|improve this answer














          The correct inequality is $(m+1)|a_{m+1}|+(m+2)|a_{m+2}|+cdots+n|a_n|<epsilon$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 18 at 23:39









          Mason

          1,9591530




          1,9591530










          answered Nov 18 at 23:10









          Kavi Rama Murthy

          50.1k31854




          50.1k31854























              2














              The series $sum_{k=0}^infty b_k$ converges absolutely $iff$ The series $sum_{k=0}^infty |b_k|$ converges.



              In your case, take $b_k = ka_k$ and plug it into whatever your definition is for (series) convergence.



              Usually, the definition of (series) convergence is the following: The series $sum_{k=0}^infty b_k$ converges to a number $L$ $iff$ "For all $epsilon > 0$, there exists an $N > 0$ such that $n geq N$ implies $left| sum_{k=1}^n b_k - L right| < epsilon$."



              As Henning Makholm correctly said in the comments, the definition you gave is for a series being Cauchy. In the case where the terms of your series are all real numbers, a series being Cauchy is equivalent to a series being convergent.






              share|cite|improve this answer




























                2














                The series $sum_{k=0}^infty b_k$ converges absolutely $iff$ The series $sum_{k=0}^infty |b_k|$ converges.



                In your case, take $b_k = ka_k$ and plug it into whatever your definition is for (series) convergence.



                Usually, the definition of (series) convergence is the following: The series $sum_{k=0}^infty b_k$ converges to a number $L$ $iff$ "For all $epsilon > 0$, there exists an $N > 0$ such that $n geq N$ implies $left| sum_{k=1}^n b_k - L right| < epsilon$."



                As Henning Makholm correctly said in the comments, the definition you gave is for a series being Cauchy. In the case where the terms of your series are all real numbers, a series being Cauchy is equivalent to a series being convergent.






                share|cite|improve this answer


























                  2












                  2








                  2






                  The series $sum_{k=0}^infty b_k$ converges absolutely $iff$ The series $sum_{k=0}^infty |b_k|$ converges.



                  In your case, take $b_k = ka_k$ and plug it into whatever your definition is for (series) convergence.



                  Usually, the definition of (series) convergence is the following: The series $sum_{k=0}^infty b_k$ converges to a number $L$ $iff$ "For all $epsilon > 0$, there exists an $N > 0$ such that $n geq N$ implies $left| sum_{k=1}^n b_k - L right| < epsilon$."



                  As Henning Makholm correctly said in the comments, the definition you gave is for a series being Cauchy. In the case where the terms of your series are all real numbers, a series being Cauchy is equivalent to a series being convergent.






                  share|cite|improve this answer














                  The series $sum_{k=0}^infty b_k$ converges absolutely $iff$ The series $sum_{k=0}^infty |b_k|$ converges.



                  In your case, take $b_k = ka_k$ and plug it into whatever your definition is for (series) convergence.



                  Usually, the definition of (series) convergence is the following: The series $sum_{k=0}^infty b_k$ converges to a number $L$ $iff$ "For all $epsilon > 0$, there exists an $N > 0$ such that $n geq N$ implies $left| sum_{k=1}^n b_k - L right| < epsilon$."



                  As Henning Makholm correctly said in the comments, the definition you gave is for a series being Cauchy. In the case where the terms of your series are all real numbers, a series being Cauchy is equivalent to a series being convergent.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 27 at 21:50

























                  answered Nov 18 at 23:26









                  Jesse Madnick

                  19.5k562123




                  19.5k562123






























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