Determine whether the sequence $sinfrac{npi}{2}$ converges and prove your answer using the epsilon method..












1














I am having the most trouble with the scratch work ahead of the actual proof. I can't figure out proof of $x_n=sinfrac{npi}{2}$ beyond $epsilon in mathbb{R}$ and $epsilon > 0$.






sequences-and-series limits epsilon-delta







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  • Have you tried computing $x_n$ for small $n$, say $0 leq n leq 5$?
    – JavaMan
    Nov 29 '18 at 3:44






  • 1




    The subsequence $x_{2n}equiv 0$. The subsequence $x_{4n+1}=1$. The subsequence $x_{4n+3}=-1$. Conclude.
    – Mark Viola
    Nov 29 '18 at 3:47


















1














I am having the most trouble with the scratch work ahead of the actual proof. I can't figure out proof of $x_n=sinfrac{npi}{2}$ beyond $epsilon in mathbb{R}$ and $epsilon > 0$.






sequences-and-series limits epsilon-delta







share|cite|improve this question
























  • Have you tried computing $x_n$ for small $n$, say $0 leq n leq 5$?
    – JavaMan
    Nov 29 '18 at 3:44






  • 1




    The subsequence $x_{2n}equiv 0$. The subsequence $x_{4n+1}=1$. The subsequence $x_{4n+3}=-1$. Conclude.
    – Mark Viola
    Nov 29 '18 at 3:47
















1












1








1







I am having the most trouble with the scratch work ahead of the actual proof. I can't figure out proof of $x_n=sinfrac{npi}{2}$ beyond $epsilon in mathbb{R}$ and $epsilon > 0$.






sequences-and-series limits epsilon-delta







share|cite|improve this question















I am having the most trouble with the scratch work ahead of the actual proof. I can't figure out proof of $x_n=sinfrac{npi}{2}$ beyond $epsilon in mathbb{R}$ and $epsilon > 0$.






sequences-and-series limits epsilon-delta




sequences-and-series limits epsilon-delta






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share|cite|improve this question













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edited Nov 29 '18 at 4:24









Key Flex

7,75741232




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asked Nov 29 '18 at 3:40









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  • Have you tried computing $x_n$ for small $n$, say $0 leq n leq 5$?
    – JavaMan
    Nov 29 '18 at 3:44






  • 1




    The subsequence $x_{2n}equiv 0$. The subsequence $x_{4n+1}=1$. The subsequence $x_{4n+3}=-1$. Conclude.
    – Mark Viola
    Nov 29 '18 at 3:47




















  • Have you tried computing $x_n$ for small $n$, say $0 leq n leq 5$?
    – JavaMan
    Nov 29 '18 at 3:44






  • 1




    The subsequence $x_{2n}equiv 0$. The subsequence $x_{4n+1}=1$. The subsequence $x_{4n+3}=-1$. Conclude.
    – Mark Viola
    Nov 29 '18 at 3:47


















Have you tried computing $x_n$ for small $n$, say $0 leq n leq 5$?
– JavaMan
Nov 29 '18 at 3:44




Have you tried computing $x_n$ for small $n$, say $0 leq n leq 5$?
– JavaMan
Nov 29 '18 at 3:44




1




1




The subsequence $x_{2n}equiv 0$. The subsequence $x_{4n+1}=1$. The subsequence $x_{4n+3}=-1$. Conclude.
– Mark Viola
Nov 29 '18 at 3:47






The subsequence $x_{2n}equiv 0$. The subsequence $x_{4n+1}=1$. The subsequence $x_{4n+3}=-1$. Conclude.
– Mark Viola
Nov 29 '18 at 3:47












4 Answers
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2














HINT:



Note that the subsequence $x_{2n}equiv 0$, the subsequence $x_{4n+1}equiv1$, and the subsequence $x_{4n+3}equiv-1$.



What can you conclude now?






share|cite|improve this answer





























    1














    It might be helpful to consider the actual values of $sin(npi / 2)$ where $n in mathbb{N}$ to obtain an alternate definition of the sequence.



    For example, try $n = 1, 2, 3, 4,...$ and so on, and see if you can find a pattern. This seems more like a trick question because of that: once you get the other, equivalent sequence, things should be much easier.



    (You might have to properly justify that the two sequences are equivalent though but that should be easy enough.)






    share|cite|improve this answer





























      1














      If the series converges to some real number $l$, then $forallvarepsilon>0, exists n_0inmathbb N$ such that $|sin frac{npi}{2}-l|<varepsilon, forall ngeq n_0$. It suffices to show that $existsvarepsilon>0$ for which no such $n_0$ exists.



      Say $varepsilon=1/10$. The interval $(l-varepsilon, l+varepsilon)equiv(l-1/10, l+1/10)$ has length 1/5. Say $sin frac{npi}{2}in(l-1/10, l+1/10)$



      $sin frac{(n+2)pi}{2}=-sinfrac{npi}{2}in(l-1/10, l+1/10)$



      or, by the definition of an interval, $(-sinfrac{npi}{2}, sin frac{npi}{2})subset(l-1/10, l+1/10)$. Whatever be $n_0$ in this case, you can always find a number of the form $n=4m+3geq n_0, minmathbb N$.



      $(-sinfrac{npi}{2}, sin frac{npi}{2})equiv(-1,1)$



      What is the length of this interval?






      share|cite|improve this answer





























        0














        Let $$a_n=x_{2n}=sin npi =0\b_n=x_{{4n+1}}=sin {4n+1over 2}pi=1$$since $a_n$ and $b_n$ are convergent to different limits, therefore $x_n$ is not convergent.






        share|cite|improve this answer





















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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2














          HINT:



          Note that the subsequence $x_{2n}equiv 0$, the subsequence $x_{4n+1}equiv1$, and the subsequence $x_{4n+3}equiv-1$.



          What can you conclude now?






          share|cite|improve this answer


























            2














            HINT:



            Note that the subsequence $x_{2n}equiv 0$, the subsequence $x_{4n+1}equiv1$, and the subsequence $x_{4n+3}equiv-1$.



            What can you conclude now?






            share|cite|improve this answer
























              2












              2








              2






              HINT:



              Note that the subsequence $x_{2n}equiv 0$, the subsequence $x_{4n+1}equiv1$, and the subsequence $x_{4n+3}equiv-1$.



              What can you conclude now?






              share|cite|improve this answer












              HINT:



              Note that the subsequence $x_{2n}equiv 0$, the subsequence $x_{4n+1}equiv1$, and the subsequence $x_{4n+3}equiv-1$.



              What can you conclude now?







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 29 '18 at 3:52









              Mark ViolaMark Viola

              130k1274170




              130k1274170























                  1














                  It might be helpful to consider the actual values of $sin(npi / 2)$ where $n in mathbb{N}$ to obtain an alternate definition of the sequence.



                  For example, try $n = 1, 2, 3, 4,...$ and so on, and see if you can find a pattern. This seems more like a trick question because of that: once you get the other, equivalent sequence, things should be much easier.



                  (You might have to properly justify that the two sequences are equivalent though but that should be easy enough.)






                  share|cite|improve this answer


























                    1














                    It might be helpful to consider the actual values of $sin(npi / 2)$ where $n in mathbb{N}$ to obtain an alternate definition of the sequence.



                    For example, try $n = 1, 2, 3, 4,...$ and so on, and see if you can find a pattern. This seems more like a trick question because of that: once you get the other, equivalent sequence, things should be much easier.



                    (You might have to properly justify that the two sequences are equivalent though but that should be easy enough.)






                    share|cite|improve this answer
























                      1












                      1








                      1






                      It might be helpful to consider the actual values of $sin(npi / 2)$ where $n in mathbb{N}$ to obtain an alternate definition of the sequence.



                      For example, try $n = 1, 2, 3, 4,...$ and so on, and see if you can find a pattern. This seems more like a trick question because of that: once you get the other, equivalent sequence, things should be much easier.



                      (You might have to properly justify that the two sequences are equivalent though but that should be easy enough.)






                      share|cite|improve this answer












                      It might be helpful to consider the actual values of $sin(npi / 2)$ where $n in mathbb{N}$ to obtain an alternate definition of the sequence.



                      For example, try $n = 1, 2, 3, 4,...$ and so on, and see if you can find a pattern. This seems more like a trick question because of that: once you get the other, equivalent sequence, things should be much easier.



                      (You might have to properly justify that the two sequences are equivalent though but that should be easy enough.)







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Nov 29 '18 at 3:44









                      Eevee TrainerEevee Trainer

                      5,1071734




                      5,1071734























                          1














                          If the series converges to some real number $l$, then $forallvarepsilon>0, exists n_0inmathbb N$ such that $|sin frac{npi}{2}-l|<varepsilon, forall ngeq n_0$. It suffices to show that $existsvarepsilon>0$ for which no such $n_0$ exists.



                          Say $varepsilon=1/10$. The interval $(l-varepsilon, l+varepsilon)equiv(l-1/10, l+1/10)$ has length 1/5. Say $sin frac{npi}{2}in(l-1/10, l+1/10)$



                          $sin frac{(n+2)pi}{2}=-sinfrac{npi}{2}in(l-1/10, l+1/10)$



                          or, by the definition of an interval, $(-sinfrac{npi}{2}, sin frac{npi}{2})subset(l-1/10, l+1/10)$. Whatever be $n_0$ in this case, you can always find a number of the form $n=4m+3geq n_0, minmathbb N$.



                          $(-sinfrac{npi}{2}, sin frac{npi}{2})equiv(-1,1)$



                          What is the length of this interval?






                          share|cite|improve this answer


























                            1














                            If the series converges to some real number $l$, then $forallvarepsilon>0, exists n_0inmathbb N$ such that $|sin frac{npi}{2}-l|<varepsilon, forall ngeq n_0$. It suffices to show that $existsvarepsilon>0$ for which no such $n_0$ exists.



                            Say $varepsilon=1/10$. The interval $(l-varepsilon, l+varepsilon)equiv(l-1/10, l+1/10)$ has length 1/5. Say $sin frac{npi}{2}in(l-1/10, l+1/10)$



                            $sin frac{(n+2)pi}{2}=-sinfrac{npi}{2}in(l-1/10, l+1/10)$



                            or, by the definition of an interval, $(-sinfrac{npi}{2}, sin frac{npi}{2})subset(l-1/10, l+1/10)$. Whatever be $n_0$ in this case, you can always find a number of the form $n=4m+3geq n_0, minmathbb N$.



                            $(-sinfrac{npi}{2}, sin frac{npi}{2})equiv(-1,1)$



                            What is the length of this interval?






                            share|cite|improve this answer
























                              1












                              1








                              1






                              If the series converges to some real number $l$, then $forallvarepsilon>0, exists n_0inmathbb N$ such that $|sin frac{npi}{2}-l|<varepsilon, forall ngeq n_0$. It suffices to show that $existsvarepsilon>0$ for which no such $n_0$ exists.



                              Say $varepsilon=1/10$. The interval $(l-varepsilon, l+varepsilon)equiv(l-1/10, l+1/10)$ has length 1/5. Say $sin frac{npi}{2}in(l-1/10, l+1/10)$



                              $sin frac{(n+2)pi}{2}=-sinfrac{npi}{2}in(l-1/10, l+1/10)$



                              or, by the definition of an interval, $(-sinfrac{npi}{2}, sin frac{npi}{2})subset(l-1/10, l+1/10)$. Whatever be $n_0$ in this case, you can always find a number of the form $n=4m+3geq n_0, minmathbb N$.



                              $(-sinfrac{npi}{2}, sin frac{npi}{2})equiv(-1,1)$



                              What is the length of this interval?






                              share|cite|improve this answer












                              If the series converges to some real number $l$, then $forallvarepsilon>0, exists n_0inmathbb N$ such that $|sin frac{npi}{2}-l|<varepsilon, forall ngeq n_0$. It suffices to show that $existsvarepsilon>0$ for which no such $n_0$ exists.



                              Say $varepsilon=1/10$. The interval $(l-varepsilon, l+varepsilon)equiv(l-1/10, l+1/10)$ has length 1/5. Say $sin frac{npi}{2}in(l-1/10, l+1/10)$



                              $sin frac{(n+2)pi}{2}=-sinfrac{npi}{2}in(l-1/10, l+1/10)$



                              or, by the definition of an interval, $(-sinfrac{npi}{2}, sin frac{npi}{2})subset(l-1/10, l+1/10)$. Whatever be $n_0$ in this case, you can always find a number of the form $n=4m+3geq n_0, minmathbb N$.



                              $(-sinfrac{npi}{2}, sin frac{npi}{2})equiv(-1,1)$



                              What is the length of this interval?







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Nov 29 '18 at 4:17









                              Shubham JohriShubham Johri

                              4,479717




                              4,479717























                                  0














                                  Let $$a_n=x_{2n}=sin npi =0\b_n=x_{{4n+1}}=sin {4n+1over 2}pi=1$$since $a_n$ and $b_n$ are convergent to different limits, therefore $x_n$ is not convergent.






                                  share|cite|improve this answer


























                                    0














                                    Let $$a_n=x_{2n}=sin npi =0\b_n=x_{{4n+1}}=sin {4n+1over 2}pi=1$$since $a_n$ and $b_n$ are convergent to different limits, therefore $x_n$ is not convergent.






                                    share|cite|improve this answer
























                                      0












                                      0








                                      0






                                      Let $$a_n=x_{2n}=sin npi =0\b_n=x_{{4n+1}}=sin {4n+1over 2}pi=1$$since $a_n$ and $b_n$ are convergent to different limits, therefore $x_n$ is not convergent.






                                      share|cite|improve this answer












                                      Let $$a_n=x_{2n}=sin npi =0\b_n=x_{{4n+1}}=sin {4n+1over 2}pi=1$$since $a_n$ and $b_n$ are convergent to different limits, therefore $x_n$ is not convergent.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Nov 29 '18 at 8:16









                                      Mostafa AyazMostafa Ayaz

                                      14.2k3937




                                      14.2k3937






























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