Question on the limit of a sequence if the sum of the sequence converges












2












$begingroup$


Suppose I have a sequence ${a_i}_{i=1}^{infty}$. I know that $a_i geq 0$ for all $i$ and $sum_{i=0}^{infty}a_i < infty$. Can I say that



$$ lim_{i to infty} a_i = 0$$



Intuitively, I think this is true. If the limit tends to a positive number, the sum would explode as well. However, I'm having a hard time proving it formally. Can I get a hint?










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    2












    $begingroup$


    Suppose I have a sequence ${a_i}_{i=1}^{infty}$. I know that $a_i geq 0$ for all $i$ and $sum_{i=0}^{infty}a_i < infty$. Can I say that



    $$ lim_{i to infty} a_i = 0$$



    Intuitively, I think this is true. If the limit tends to a positive number, the sum would explode as well. However, I'm having a hard time proving it formally. Can I get a hint?










    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      Suppose I have a sequence ${a_i}_{i=1}^{infty}$. I know that $a_i geq 0$ for all $i$ and $sum_{i=0}^{infty}a_i < infty$. Can I say that



      $$ lim_{i to infty} a_i = 0$$



      Intuitively, I think this is true. If the limit tends to a positive number, the sum would explode as well. However, I'm having a hard time proving it formally. Can I get a hint?










      share|cite|improve this question









      $endgroup$




      Suppose I have a sequence ${a_i}_{i=1}^{infty}$. I know that $a_i geq 0$ for all $i$ and $sum_{i=0}^{infty}a_i < infty$. Can I say that



      $$ lim_{i to infty} a_i = 0$$



      Intuitively, I think this is true. If the limit tends to a positive number, the sum would explode as well. However, I'm having a hard time proving it formally. Can I get a hint?







      sequences-and-series limits summation






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      asked 6 hours ago









      user1691278user1691278

      47139




      47139






















          3 Answers
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          $begingroup$

          We have
          $$sum_{i=0}^n a_ito S ,qquad sum_{i=0}^{n-1} a_ito S$$
          as $ntoinfty$, so
          $$a_n=Bigl(sum_{i=0}^n a_iBigr)-Bigl(sum_{i=0}^{n-1} a_iBigr)to S-S=0 .$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Great proof. Is $a_i > 0$ even necessary?
            $endgroup$
            – user1691278
            6 hours ago












          • $begingroup$
            No.$!,!,!,$
            $endgroup$
            – David
            6 hours ago



















          1












          $begingroup$

          I'm not quite sure if the sequence starts at index $0$ or $1$ as your question uses $1$ as the sequence start but the summation starts at $0$. As it's not important, I will assume it's $0$. Note that



          $$sum_{i , = , 0}^{infty} a_i lt infty tag{1}label{eq1}$$



          usually means it converges to a limit. However, consider the somewhat looser restriction of there just simply existing a real supremum, call it $S$, so that



          $$sum_{i , = , 0}^{n} a_i le S ; forall ; n ge 0 tag{2}label{eq2}$$



          As trying to show the limit of $a_i$ is $0$, I will just let the $- 0$ be implied. Also, it's given that $a_i ge 0$, so the partial sums are non-decreasing, and there are no non-negative values so I don't need to use absolute values in the following. For proving that



          $$lim_{i , to , infty} a_i = 0 tag{3}label{eq3}$$



          consider that for any $epsilon gt 0$ there must be a finite number of integers for which $a_n ge epsilon$ as, otherwise, the sum of them would be unbounded. Thus, there must be either none or a maximum index, but in either choose an integer $n_0$ to be greater than any such value and, thus, $a_n lt epsilon$ for all $n gt n_0$, which confirms that eqref{eq3} is true. I won't show it here, but the infinite sum has a limit as well of $S$.



          Note that assuming just a supremum for the sums without assuming that $a_i ge 0$ won't suffice to show the limit as, for example, $a_i = left(-1right)^i$ would cause the partial sums to alternate between $1$ and $0$, thus allowing a supremum of $1$, but with $a_i$ not converging to $0$.



          Note that requiring the sum of $a_i$ to have a finite limit means that, as David indicated in his answer's comment, also requiring the $a_i$ to be non-negative is unnecessary. However, if the requirement of eqref{eq1} includes the possibility of the limit being $-infty$, then some appropriate restrictions on $a_i$ would be required.






          share|cite|improve this answer











          $endgroup$





















            0












            $begingroup$

            Informal outline for a more formal proof:



            Let $s_n$ be the sum of the first $n$ terms and let the limit of the series be $m$.



            Convergence means that for as small an $epsilon>0$ as we like and large enough $N$, $n>N$ guarantees $|m-s_n|<epsilon$.



            But this goes awry if ${a_n}$ doesn't converge to $0$, because when $2epsilon<|a_{n+1}|$ it's not possible to have both $|m-s_n|<epsilon$ and $|m-s_{n+1}|<epsilon$.



            (And use the definition of ${a_n}$ not converging to show that this situation must arise.)






            share|cite|improve this answer











            $endgroup$













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              3 Answers
              3






              active

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              3 Answers
              3






              active

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              5












              $begingroup$

              We have
              $$sum_{i=0}^n a_ito S ,qquad sum_{i=0}^{n-1} a_ito S$$
              as $ntoinfty$, so
              $$a_n=Bigl(sum_{i=0}^n a_iBigr)-Bigl(sum_{i=0}^{n-1} a_iBigr)to S-S=0 .$$






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Great proof. Is $a_i > 0$ even necessary?
                $endgroup$
                – user1691278
                6 hours ago












              • $begingroup$
                No.$!,!,!,$
                $endgroup$
                – David
                6 hours ago
















              5












              $begingroup$

              We have
              $$sum_{i=0}^n a_ito S ,qquad sum_{i=0}^{n-1} a_ito S$$
              as $ntoinfty$, so
              $$a_n=Bigl(sum_{i=0}^n a_iBigr)-Bigl(sum_{i=0}^{n-1} a_iBigr)to S-S=0 .$$






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Great proof. Is $a_i > 0$ even necessary?
                $endgroup$
                – user1691278
                6 hours ago












              • $begingroup$
                No.$!,!,!,$
                $endgroup$
                – David
                6 hours ago














              5












              5








              5





              $begingroup$

              We have
              $$sum_{i=0}^n a_ito S ,qquad sum_{i=0}^{n-1} a_ito S$$
              as $ntoinfty$, so
              $$a_n=Bigl(sum_{i=0}^n a_iBigr)-Bigl(sum_{i=0}^{n-1} a_iBigr)to S-S=0 .$$






              share|cite|improve this answer









              $endgroup$



              We have
              $$sum_{i=0}^n a_ito S ,qquad sum_{i=0}^{n-1} a_ito S$$
              as $ntoinfty$, so
              $$a_n=Bigl(sum_{i=0}^n a_iBigr)-Bigl(sum_{i=0}^{n-1} a_iBigr)to S-S=0 .$$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered 6 hours ago









              DavidDavid

              68k664126




              68k664126












              • $begingroup$
                Great proof. Is $a_i > 0$ even necessary?
                $endgroup$
                – user1691278
                6 hours ago












              • $begingroup$
                No.$!,!,!,$
                $endgroup$
                – David
                6 hours ago


















              • $begingroup$
                Great proof. Is $a_i > 0$ even necessary?
                $endgroup$
                – user1691278
                6 hours ago












              • $begingroup$
                No.$!,!,!,$
                $endgroup$
                – David
                6 hours ago
















              $begingroup$
              Great proof. Is $a_i > 0$ even necessary?
              $endgroup$
              – user1691278
              6 hours ago






              $begingroup$
              Great proof. Is $a_i > 0$ even necessary?
              $endgroup$
              – user1691278
              6 hours ago














              $begingroup$
              No.$!,!,!,$
              $endgroup$
              – David
              6 hours ago




              $begingroup$
              No.$!,!,!,$
              $endgroup$
              – David
              6 hours ago











              1












              $begingroup$

              I'm not quite sure if the sequence starts at index $0$ or $1$ as your question uses $1$ as the sequence start but the summation starts at $0$. As it's not important, I will assume it's $0$. Note that



              $$sum_{i , = , 0}^{infty} a_i lt infty tag{1}label{eq1}$$



              usually means it converges to a limit. However, consider the somewhat looser restriction of there just simply existing a real supremum, call it $S$, so that



              $$sum_{i , = , 0}^{n} a_i le S ; forall ; n ge 0 tag{2}label{eq2}$$



              As trying to show the limit of $a_i$ is $0$, I will just let the $- 0$ be implied. Also, it's given that $a_i ge 0$, so the partial sums are non-decreasing, and there are no non-negative values so I don't need to use absolute values in the following. For proving that



              $$lim_{i , to , infty} a_i = 0 tag{3}label{eq3}$$



              consider that for any $epsilon gt 0$ there must be a finite number of integers for which $a_n ge epsilon$ as, otherwise, the sum of them would be unbounded. Thus, there must be either none or a maximum index, but in either choose an integer $n_0$ to be greater than any such value and, thus, $a_n lt epsilon$ for all $n gt n_0$, which confirms that eqref{eq3} is true. I won't show it here, but the infinite sum has a limit as well of $S$.



              Note that assuming just a supremum for the sums without assuming that $a_i ge 0$ won't suffice to show the limit as, for example, $a_i = left(-1right)^i$ would cause the partial sums to alternate between $1$ and $0$, thus allowing a supremum of $1$, but with $a_i$ not converging to $0$.



              Note that requiring the sum of $a_i$ to have a finite limit means that, as David indicated in his answer's comment, also requiring the $a_i$ to be non-negative is unnecessary. However, if the requirement of eqref{eq1} includes the possibility of the limit being $-infty$, then some appropriate restrictions on $a_i$ would be required.






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                I'm not quite sure if the sequence starts at index $0$ or $1$ as your question uses $1$ as the sequence start but the summation starts at $0$. As it's not important, I will assume it's $0$. Note that



                $$sum_{i , = , 0}^{infty} a_i lt infty tag{1}label{eq1}$$



                usually means it converges to a limit. However, consider the somewhat looser restriction of there just simply existing a real supremum, call it $S$, so that



                $$sum_{i , = , 0}^{n} a_i le S ; forall ; n ge 0 tag{2}label{eq2}$$



                As trying to show the limit of $a_i$ is $0$, I will just let the $- 0$ be implied. Also, it's given that $a_i ge 0$, so the partial sums are non-decreasing, and there are no non-negative values so I don't need to use absolute values in the following. For proving that



                $$lim_{i , to , infty} a_i = 0 tag{3}label{eq3}$$



                consider that for any $epsilon gt 0$ there must be a finite number of integers for which $a_n ge epsilon$ as, otherwise, the sum of them would be unbounded. Thus, there must be either none or a maximum index, but in either choose an integer $n_0$ to be greater than any such value and, thus, $a_n lt epsilon$ for all $n gt n_0$, which confirms that eqref{eq3} is true. I won't show it here, but the infinite sum has a limit as well of $S$.



                Note that assuming just a supremum for the sums without assuming that $a_i ge 0$ won't suffice to show the limit as, for example, $a_i = left(-1right)^i$ would cause the partial sums to alternate between $1$ and $0$, thus allowing a supremum of $1$, but with $a_i$ not converging to $0$.



                Note that requiring the sum of $a_i$ to have a finite limit means that, as David indicated in his answer's comment, also requiring the $a_i$ to be non-negative is unnecessary. However, if the requirement of eqref{eq1} includes the possibility of the limit being $-infty$, then some appropriate restrictions on $a_i$ would be required.






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  I'm not quite sure if the sequence starts at index $0$ or $1$ as your question uses $1$ as the sequence start but the summation starts at $0$. As it's not important, I will assume it's $0$. Note that



                  $$sum_{i , = , 0}^{infty} a_i lt infty tag{1}label{eq1}$$



                  usually means it converges to a limit. However, consider the somewhat looser restriction of there just simply existing a real supremum, call it $S$, so that



                  $$sum_{i , = , 0}^{n} a_i le S ; forall ; n ge 0 tag{2}label{eq2}$$



                  As trying to show the limit of $a_i$ is $0$, I will just let the $- 0$ be implied. Also, it's given that $a_i ge 0$, so the partial sums are non-decreasing, and there are no non-negative values so I don't need to use absolute values in the following. For proving that



                  $$lim_{i , to , infty} a_i = 0 tag{3}label{eq3}$$



                  consider that for any $epsilon gt 0$ there must be a finite number of integers for which $a_n ge epsilon$ as, otherwise, the sum of them would be unbounded. Thus, there must be either none or a maximum index, but in either choose an integer $n_0$ to be greater than any such value and, thus, $a_n lt epsilon$ for all $n gt n_0$, which confirms that eqref{eq3} is true. I won't show it here, but the infinite sum has a limit as well of $S$.



                  Note that assuming just a supremum for the sums without assuming that $a_i ge 0$ won't suffice to show the limit as, for example, $a_i = left(-1right)^i$ would cause the partial sums to alternate between $1$ and $0$, thus allowing a supremum of $1$, but with $a_i$ not converging to $0$.



                  Note that requiring the sum of $a_i$ to have a finite limit means that, as David indicated in his answer's comment, also requiring the $a_i$ to be non-negative is unnecessary. However, if the requirement of eqref{eq1} includes the possibility of the limit being $-infty$, then some appropriate restrictions on $a_i$ would be required.






                  share|cite|improve this answer











                  $endgroup$



                  I'm not quite sure if the sequence starts at index $0$ or $1$ as your question uses $1$ as the sequence start but the summation starts at $0$. As it's not important, I will assume it's $0$. Note that



                  $$sum_{i , = , 0}^{infty} a_i lt infty tag{1}label{eq1}$$



                  usually means it converges to a limit. However, consider the somewhat looser restriction of there just simply existing a real supremum, call it $S$, so that



                  $$sum_{i , = , 0}^{n} a_i le S ; forall ; n ge 0 tag{2}label{eq2}$$



                  As trying to show the limit of $a_i$ is $0$, I will just let the $- 0$ be implied. Also, it's given that $a_i ge 0$, so the partial sums are non-decreasing, and there are no non-negative values so I don't need to use absolute values in the following. For proving that



                  $$lim_{i , to , infty} a_i = 0 tag{3}label{eq3}$$



                  consider that for any $epsilon gt 0$ there must be a finite number of integers for which $a_n ge epsilon$ as, otherwise, the sum of them would be unbounded. Thus, there must be either none or a maximum index, but in either choose an integer $n_0$ to be greater than any such value and, thus, $a_n lt epsilon$ for all $n gt n_0$, which confirms that eqref{eq3} is true. I won't show it here, but the infinite sum has a limit as well of $S$.



                  Note that assuming just a supremum for the sums without assuming that $a_i ge 0$ won't suffice to show the limit as, for example, $a_i = left(-1right)^i$ would cause the partial sums to alternate between $1$ and $0$, thus allowing a supremum of $1$, but with $a_i$ not converging to $0$.



                  Note that requiring the sum of $a_i$ to have a finite limit means that, as David indicated in his answer's comment, also requiring the $a_i$ to be non-negative is unnecessary. However, if the requirement of eqref{eq1} includes the possibility of the limit being $-infty$, then some appropriate restrictions on $a_i$ would be required.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 4 hours ago

























                  answered 5 hours ago









                  John OmielanJohn Omielan

                  1,28629




                  1,28629























                      0












                      $begingroup$

                      Informal outline for a more formal proof:



                      Let $s_n$ be the sum of the first $n$ terms and let the limit of the series be $m$.



                      Convergence means that for as small an $epsilon>0$ as we like and large enough $N$, $n>N$ guarantees $|m-s_n|<epsilon$.



                      But this goes awry if ${a_n}$ doesn't converge to $0$, because when $2epsilon<|a_{n+1}|$ it's not possible to have both $|m-s_n|<epsilon$ and $|m-s_{n+1}|<epsilon$.



                      (And use the definition of ${a_n}$ not converging to show that this situation must arise.)






                      share|cite|improve this answer











                      $endgroup$


















                        0












                        $begingroup$

                        Informal outline for a more formal proof:



                        Let $s_n$ be the sum of the first $n$ terms and let the limit of the series be $m$.



                        Convergence means that for as small an $epsilon>0$ as we like and large enough $N$, $n>N$ guarantees $|m-s_n|<epsilon$.



                        But this goes awry if ${a_n}$ doesn't converge to $0$, because when $2epsilon<|a_{n+1}|$ it's not possible to have both $|m-s_n|<epsilon$ and $|m-s_{n+1}|<epsilon$.



                        (And use the definition of ${a_n}$ not converging to show that this situation must arise.)






                        share|cite|improve this answer











                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Informal outline for a more formal proof:



                          Let $s_n$ be the sum of the first $n$ terms and let the limit of the series be $m$.



                          Convergence means that for as small an $epsilon>0$ as we like and large enough $N$, $n>N$ guarantees $|m-s_n|<epsilon$.



                          But this goes awry if ${a_n}$ doesn't converge to $0$, because when $2epsilon<|a_{n+1}|$ it's not possible to have both $|m-s_n|<epsilon$ and $|m-s_{n+1}|<epsilon$.



                          (And use the definition of ${a_n}$ not converging to show that this situation must arise.)






                          share|cite|improve this answer











                          $endgroup$



                          Informal outline for a more formal proof:



                          Let $s_n$ be the sum of the first $n$ terms and let the limit of the series be $m$.



                          Convergence means that for as small an $epsilon>0$ as we like and large enough $N$, $n>N$ guarantees $|m-s_n|<epsilon$.



                          But this goes awry if ${a_n}$ doesn't converge to $0$, because when $2epsilon<|a_{n+1}|$ it's not possible to have both $|m-s_n|<epsilon$ and $|m-s_{n+1}|<epsilon$.



                          (And use the definition of ${a_n}$ not converging to show that this situation must arise.)







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited 5 hours ago

























                          answered 5 hours ago









                          timtfjtimtfj

                          1,318318




                          1,318318






























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