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











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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 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 at 3:47

















up vote
1
down vote

favorite












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 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 at 3:47















up vote
1
down vote

favorite









up vote
1
down vote

favorite











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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edited Nov 29 at 4:24









Key Flex

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asked Nov 29 at 3:40









NeedHelp

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  • Have you tried computing $x_n$ for small $n$, say $0 leq n leq 5$?
    – JavaMan
    Nov 29 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 at 3:47




















  • Have you tried computing $x_n$ for small $n$, say $0 leq n leq 5$?
    – JavaMan
    Nov 29 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 at 3:47


















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




Have you tried computing $x_n$ for small $n$, say $0 leq n leq 5$?
– JavaMan
Nov 29 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 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 at 3:47












4 Answers
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2
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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




























    up vote
    1
    down vote













    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




























      up vote
      1
      down vote













      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




























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        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.






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          4 Answers
          4






          active

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          4 Answers
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          up vote
          2
          down vote













          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

























            up vote
            2
            down vote













            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























              up vote
              2
              down vote










              up vote
              2
              down vote









              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 at 3:52









              Mark Viola

              130k1273170




              130k1273170






















                  up vote
                  1
                  down vote













                  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

























                    up vote
                    1
                    down vote













                    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























                      up vote
                      1
                      down vote










                      up vote
                      1
                      down vote









                      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 at 3:44









                      Eevee Trainer

                      3,370225




                      3,370225






















                          up vote
                          1
                          down vote













                          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

























                            up vote
                            1
                            down vote













                            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























                              up vote
                              1
                              down vote










                              up vote
                              1
                              down vote









                              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 at 4:17









                              Shubham Johri

                              3,084413




                              3,084413






















                                  up vote
                                  0
                                  down vote













                                  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

























                                    up vote
                                    0
                                    down vote













                                    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























                                      up vote
                                      0
                                      down vote










                                      up vote
                                      0
                                      down vote









                                      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 at 8:16









                                      Mostafa Ayaz

                                      13.6k3836




                                      13.6k3836






























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