How would I determine when $sin(100) < 1$ to test for convergency of this series?












0












$begingroup$


My Stewart calculus textbook gives the following problem and asks to determine if the series converges or diverges, and to find the sum if it converges: $sum_{k=1}^infty (sin(100))^k$



Now, I know that this is a geometric series and that it will converge for when $|r| < 1$. But what I don't understand is how to determine whether $sin(100) < 1$ without using a calculator.






sequences-and-series convergence







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    $|sin(x)| = 1$ only when $x = pmfrac{pi}{2},pmfrac{3pi}{2},pmfrac{5pi}{2},ldots$. You just need to show that $100$ cannot be written on this form.
    $endgroup$
    – Winther
    Oct 3 '16 at 22:51












  • $begingroup$
    I'm assuming 100 is in radians; the book doesn't specify
    $endgroup$
    – AleksandrH
    Oct 3 '16 at 22:51












  • $begingroup$
    It's $<1$ even if $100$ is in degrees. To show this use that $100^circ = frac{2pi}{360}cdot 100 = frac{5pi}{9}$.
    $endgroup$
    – Winther
    Oct 3 '16 at 22:55


















0












$begingroup$


My Stewart calculus textbook gives the following problem and asks to determine if the series converges or diverges, and to find the sum if it converges: $sum_{k=1}^infty (sin(100))^k$



Now, I know that this is a geometric series and that it will converge for when $|r| < 1$. But what I don't understand is how to determine whether $sin(100) < 1$ without using a calculator.






sequences-and-series convergence







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    $|sin(x)| = 1$ only when $x = pmfrac{pi}{2},pmfrac{3pi}{2},pmfrac{5pi}{2},ldots$. You just need to show that $100$ cannot be written on this form.
    $endgroup$
    – Winther
    Oct 3 '16 at 22:51












  • $begingroup$
    I'm assuming 100 is in radians; the book doesn't specify
    $endgroup$
    – AleksandrH
    Oct 3 '16 at 22:51












  • $begingroup$
    It's $<1$ even if $100$ is in degrees. To show this use that $100^circ = frac{2pi}{360}cdot 100 = frac{5pi}{9}$.
    $endgroup$
    – Winther
    Oct 3 '16 at 22:55
















0












0








0





$begingroup$


My Stewart calculus textbook gives the following problem and asks to determine if the series converges or diverges, and to find the sum if it converges: $sum_{k=1}^infty (sin(100))^k$



Now, I know that this is a geometric series and that it will converge for when $|r| < 1$. But what I don't understand is how to determine whether $sin(100) < 1$ without using a calculator.






sequences-and-series convergence







share|cite|improve this question











$endgroup$




My Stewart calculus textbook gives the following problem and asks to determine if the series converges or diverges, and to find the sum if it converges: $sum_{k=1}^infty (sin(100))^k$



Now, I know that this is a geometric series and that it will converge for when $|r| < 1$. But what I don't understand is how to determine whether $sin(100) < 1$ without using a calculator.






sequences-and-series convergence




sequences-and-series convergence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Oct 3 '16 at 22:56









Winther

20.6k33156




20.6k33156










asked Oct 3 '16 at 22:49









AleksandrHAleksandrH

1,23721123




1,23721123








  • 3




    $begingroup$
    $|sin(x)| = 1$ only when $x = pmfrac{pi}{2},pmfrac{3pi}{2},pmfrac{5pi}{2},ldots$. You just need to show that $100$ cannot be written on this form.
    $endgroup$
    – Winther
    Oct 3 '16 at 22:51












  • $begingroup$
    I'm assuming 100 is in radians; the book doesn't specify
    $endgroup$
    – AleksandrH
    Oct 3 '16 at 22:51












  • $begingroup$
    It's $<1$ even if $100$ is in degrees. To show this use that $100^circ = frac{2pi}{360}cdot 100 = frac{5pi}{9}$.
    $endgroup$
    – Winther
    Oct 3 '16 at 22:55
















  • 3




    $begingroup$
    $|sin(x)| = 1$ only when $x = pmfrac{pi}{2},pmfrac{3pi}{2},pmfrac{5pi}{2},ldots$. You just need to show that $100$ cannot be written on this form.
    $endgroup$
    – Winther
    Oct 3 '16 at 22:51












  • $begingroup$
    I'm assuming 100 is in radians; the book doesn't specify
    $endgroup$
    – AleksandrH
    Oct 3 '16 at 22:51












  • $begingroup$
    It's $<1$ even if $100$ is in degrees. To show this use that $100^circ = frac{2pi}{360}cdot 100 = frac{5pi}{9}$.
    $endgroup$
    – Winther
    Oct 3 '16 at 22:55










3




3




$begingroup$
$|sin(x)| = 1$ only when $x = pmfrac{pi}{2},pmfrac{3pi}{2},pmfrac{5pi}{2},ldots$. You just need to show that $100$ cannot be written on this form.
$endgroup$
– Winther
Oct 3 '16 at 22:51






$begingroup$
$|sin(x)| = 1$ only when $x = pmfrac{pi}{2},pmfrac{3pi}{2},pmfrac{5pi}{2},ldots$. You just need to show that $100$ cannot be written on this form.
$endgroup$
– Winther
Oct 3 '16 at 22:51














$begingroup$
I'm assuming 100 is in radians; the book doesn't specify
$endgroup$
– AleksandrH
Oct 3 '16 at 22:51






$begingroup$
I'm assuming 100 is in radians; the book doesn't specify
$endgroup$
– AleksandrH
Oct 3 '16 at 22:51














$begingroup$
It's $<1$ even if $100$ is in degrees. To show this use that $100^circ = frac{2pi}{360}cdot 100 = frac{5pi}{9}$.
$endgroup$
– Winther
Oct 3 '16 at 22:55






$begingroup$
It's $<1$ even if $100$ is in degrees. To show this use that $100^circ = frac{2pi}{360}cdot 100 = frac{5pi}{9}$.
$endgroup$
– Winther
Oct 3 '16 at 22:55












2 Answers
2






active

oldest

votes


















3












$begingroup$

$|sin (x)|=1$ if $x$ is an odd multiple of $frac{pi}{2}$



Suppose on the contradiction that $$|sin(100)|=1$$ Then we can conclude that $$100=kfrac{pi}{2}$$



where $k$ is an odd integer.



Then $pi=frac{200}{k}$ but $pi$ is irrational but $frac{200}{k}$ is rational, hence we get a contradiction. $$|sin(100)|<1.$$



and we have $$sin(100)<1$$



The above working is for $100$ being interpreted as radian.



Suppose it is in degree, well, $100$ is not a multiple of $90$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I hate that I'm this far in mathematics and that I still get tripped up by simple things like this. What's the significance of pi being irrational?
    $endgroup$
    – AleksandrH
    Oct 3 '16 at 22:53






  • 1




    $begingroup$
    The left hand side is irrational but the right hand side is rational, hence we get a contradiction, Hence our initial assumption must be wrong.
    $endgroup$
    – Siong Thye Goh
    Oct 3 '16 at 22:58






  • 1




    $begingroup$
    |sin(angle)| is equal to $1$ if and only if it is an odd multiple of $frac{pi}{2}$
    $endgroup$
    – Siong Thye Goh
    Oct 4 '16 at 1:32






  • 1




    $begingroup$
    @AleksandrH Since $pi$ is irrational, no integer is an integer multiple of $frac{pi}{2}$, so $|sin(n)| neq 1$ for integer $n$.
    $endgroup$
    – Carl Schildkraut
    Oct 4 '16 at 1:58






  • 1




    $begingroup$
    I'm so stupid, I just realized that if sin(100) is never 1, then it is always less than 1, and so it must converge. Urgh, hate calculus. Thanks.
    $endgroup$
    – AleksandrH
    Oct 4 '16 at 13:02



















1












$begingroup$

If $sin(100)=1$, then $100$ would be an integer multiple of $pi/2$.

That is,
$$
frac{200}{pi}
$$
would have to be an integer.



Calculating, we find $frac{200}{pi} approx 63.661... $ so it appears not to be an integer. Proving this requires bounds on $pi$.



If we are willing to accept that
$$
3.14 < pi < 3.142
$$
then we may conclude that
$$
63.65372374... < frac{200}{pi} < 63.694267515...
$$
and since there is no integer between these two bounds, we know $frac{200}{pi}$ is not an integer, and hence $sin(100) neq 1$.






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    $|sin (x)|=1$ if $x$ is an odd multiple of $frac{pi}{2}$



    Suppose on the contradiction that $$|sin(100)|=1$$ Then we can conclude that $$100=kfrac{pi}{2}$$



    where $k$ is an odd integer.



    Then $pi=frac{200}{k}$ but $pi$ is irrational but $frac{200}{k}$ is rational, hence we get a contradiction. $$|sin(100)|<1.$$



    and we have $$sin(100)<1$$



    The above working is for $100$ being interpreted as radian.



    Suppose it is in degree, well, $100$ is not a multiple of $90$.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      I hate that I'm this far in mathematics and that I still get tripped up by simple things like this. What's the significance of pi being irrational?
      $endgroup$
      – AleksandrH
      Oct 3 '16 at 22:53






    • 1




      $begingroup$
      The left hand side is irrational but the right hand side is rational, hence we get a contradiction, Hence our initial assumption must be wrong.
      $endgroup$
      – Siong Thye Goh
      Oct 3 '16 at 22:58






    • 1




      $begingroup$
      |sin(angle)| is equal to $1$ if and only if it is an odd multiple of $frac{pi}{2}$
      $endgroup$
      – Siong Thye Goh
      Oct 4 '16 at 1:32






    • 1




      $begingroup$
      @AleksandrH Since $pi$ is irrational, no integer is an integer multiple of $frac{pi}{2}$, so $|sin(n)| neq 1$ for integer $n$.
      $endgroup$
      – Carl Schildkraut
      Oct 4 '16 at 1:58






    • 1




      $begingroup$
      I'm so stupid, I just realized that if sin(100) is never 1, then it is always less than 1, and so it must converge. Urgh, hate calculus. Thanks.
      $endgroup$
      – AleksandrH
      Oct 4 '16 at 13:02
















    3












    $begingroup$

    $|sin (x)|=1$ if $x$ is an odd multiple of $frac{pi}{2}$



    Suppose on the contradiction that $$|sin(100)|=1$$ Then we can conclude that $$100=kfrac{pi}{2}$$



    where $k$ is an odd integer.



    Then $pi=frac{200}{k}$ but $pi$ is irrational but $frac{200}{k}$ is rational, hence we get a contradiction. $$|sin(100)|<1.$$



    and we have $$sin(100)<1$$



    The above working is for $100$ being interpreted as radian.



    Suppose it is in degree, well, $100$ is not a multiple of $90$.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      I hate that I'm this far in mathematics and that I still get tripped up by simple things like this. What's the significance of pi being irrational?
      $endgroup$
      – AleksandrH
      Oct 3 '16 at 22:53






    • 1




      $begingroup$
      The left hand side is irrational but the right hand side is rational, hence we get a contradiction, Hence our initial assumption must be wrong.
      $endgroup$
      – Siong Thye Goh
      Oct 3 '16 at 22:58






    • 1




      $begingroup$
      |sin(angle)| is equal to $1$ if and only if it is an odd multiple of $frac{pi}{2}$
      $endgroup$
      – Siong Thye Goh
      Oct 4 '16 at 1:32






    • 1




      $begingroup$
      @AleksandrH Since $pi$ is irrational, no integer is an integer multiple of $frac{pi}{2}$, so $|sin(n)| neq 1$ for integer $n$.
      $endgroup$
      – Carl Schildkraut
      Oct 4 '16 at 1:58






    • 1




      $begingroup$
      I'm so stupid, I just realized that if sin(100) is never 1, then it is always less than 1, and so it must converge. Urgh, hate calculus. Thanks.
      $endgroup$
      – AleksandrH
      Oct 4 '16 at 13:02














    3












    3








    3





    $begingroup$

    $|sin (x)|=1$ if $x$ is an odd multiple of $frac{pi}{2}$



    Suppose on the contradiction that $$|sin(100)|=1$$ Then we can conclude that $$100=kfrac{pi}{2}$$



    where $k$ is an odd integer.



    Then $pi=frac{200}{k}$ but $pi$ is irrational but $frac{200}{k}$ is rational, hence we get a contradiction. $$|sin(100)|<1.$$



    and we have $$sin(100)<1$$



    The above working is for $100$ being interpreted as radian.



    Suppose it is in degree, well, $100$ is not a multiple of $90$.






    share|cite|improve this answer











    $endgroup$



    $|sin (x)|=1$ if $x$ is an odd multiple of $frac{pi}{2}$



    Suppose on the contradiction that $$|sin(100)|=1$$ Then we can conclude that $$100=kfrac{pi}{2}$$



    where $k$ is an odd integer.



    Then $pi=frac{200}{k}$ but $pi$ is irrational but $frac{200}{k}$ is rational, hence we get a contradiction. $$|sin(100)|<1.$$



    and we have $$sin(100)<1$$



    The above working is for $100$ being interpreted as radian.



    Suppose it is in degree, well, $100$ is not a multiple of $90$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Oct 4 '16 at 1:57

























    answered Oct 3 '16 at 22:52









    Siong Thye GohSiong Thye Goh

    100k1465117




    100k1465117








    • 1




      $begingroup$
      I hate that I'm this far in mathematics and that I still get tripped up by simple things like this. What's the significance of pi being irrational?
      $endgroup$
      – AleksandrH
      Oct 3 '16 at 22:53






    • 1




      $begingroup$
      The left hand side is irrational but the right hand side is rational, hence we get a contradiction, Hence our initial assumption must be wrong.
      $endgroup$
      – Siong Thye Goh
      Oct 3 '16 at 22:58






    • 1




      $begingroup$
      |sin(angle)| is equal to $1$ if and only if it is an odd multiple of $frac{pi}{2}$
      $endgroup$
      – Siong Thye Goh
      Oct 4 '16 at 1:32






    • 1




      $begingroup$
      @AleksandrH Since $pi$ is irrational, no integer is an integer multiple of $frac{pi}{2}$, so $|sin(n)| neq 1$ for integer $n$.
      $endgroup$
      – Carl Schildkraut
      Oct 4 '16 at 1:58






    • 1




      $begingroup$
      I'm so stupid, I just realized that if sin(100) is never 1, then it is always less than 1, and so it must converge. Urgh, hate calculus. Thanks.
      $endgroup$
      – AleksandrH
      Oct 4 '16 at 13:02














    • 1




      $begingroup$
      I hate that I'm this far in mathematics and that I still get tripped up by simple things like this. What's the significance of pi being irrational?
      $endgroup$
      – AleksandrH
      Oct 3 '16 at 22:53






    • 1




      $begingroup$
      The left hand side is irrational but the right hand side is rational, hence we get a contradiction, Hence our initial assumption must be wrong.
      $endgroup$
      – Siong Thye Goh
      Oct 3 '16 at 22:58






    • 1




      $begingroup$
      |sin(angle)| is equal to $1$ if and only if it is an odd multiple of $frac{pi}{2}$
      $endgroup$
      – Siong Thye Goh
      Oct 4 '16 at 1:32






    • 1




      $begingroup$
      @AleksandrH Since $pi$ is irrational, no integer is an integer multiple of $frac{pi}{2}$, so $|sin(n)| neq 1$ for integer $n$.
      $endgroup$
      – Carl Schildkraut
      Oct 4 '16 at 1:58






    • 1




      $begingroup$
      I'm so stupid, I just realized that if sin(100) is never 1, then it is always less than 1, and so it must converge. Urgh, hate calculus. Thanks.
      $endgroup$
      – AleksandrH
      Oct 4 '16 at 13:02








    1




    1




    $begingroup$
    I hate that I'm this far in mathematics and that I still get tripped up by simple things like this. What's the significance of pi being irrational?
    $endgroup$
    – AleksandrH
    Oct 3 '16 at 22:53




    $begingroup$
    I hate that I'm this far in mathematics and that I still get tripped up by simple things like this. What's the significance of pi being irrational?
    $endgroup$
    – AleksandrH
    Oct 3 '16 at 22:53




    1




    1




    $begingroup$
    The left hand side is irrational but the right hand side is rational, hence we get a contradiction, Hence our initial assumption must be wrong.
    $endgroup$
    – Siong Thye Goh
    Oct 3 '16 at 22:58




    $begingroup$
    The left hand side is irrational but the right hand side is rational, hence we get a contradiction, Hence our initial assumption must be wrong.
    $endgroup$
    – Siong Thye Goh
    Oct 3 '16 at 22:58




    1




    1




    $begingroup$
    |sin(angle)| is equal to $1$ if and only if it is an odd multiple of $frac{pi}{2}$
    $endgroup$
    – Siong Thye Goh
    Oct 4 '16 at 1:32




    $begingroup$
    |sin(angle)| is equal to $1$ if and only if it is an odd multiple of $frac{pi}{2}$
    $endgroup$
    – Siong Thye Goh
    Oct 4 '16 at 1:32




    1




    1




    $begingroup$
    @AleksandrH Since $pi$ is irrational, no integer is an integer multiple of $frac{pi}{2}$, so $|sin(n)| neq 1$ for integer $n$.
    $endgroup$
    – Carl Schildkraut
    Oct 4 '16 at 1:58




    $begingroup$
    @AleksandrH Since $pi$ is irrational, no integer is an integer multiple of $frac{pi}{2}$, so $|sin(n)| neq 1$ for integer $n$.
    $endgroup$
    – Carl Schildkraut
    Oct 4 '16 at 1:58




    1




    1




    $begingroup$
    I'm so stupid, I just realized that if sin(100) is never 1, then it is always less than 1, and so it must converge. Urgh, hate calculus. Thanks.
    $endgroup$
    – AleksandrH
    Oct 4 '16 at 13:02




    $begingroup$
    I'm so stupid, I just realized that if sin(100) is never 1, then it is always less than 1, and so it must converge. Urgh, hate calculus. Thanks.
    $endgroup$
    – AleksandrH
    Oct 4 '16 at 13:02











    1












    $begingroup$

    If $sin(100)=1$, then $100$ would be an integer multiple of $pi/2$.

    That is,
    $$
    frac{200}{pi}
    $$
    would have to be an integer.



    Calculating, we find $frac{200}{pi} approx 63.661... $ so it appears not to be an integer. Proving this requires bounds on $pi$.



    If we are willing to accept that
    $$
    3.14 < pi < 3.142
    $$
    then we may conclude that
    $$
    63.65372374... < frac{200}{pi} < 63.694267515...
    $$
    and since there is no integer between these two bounds, we know $frac{200}{pi}$ is not an integer, and hence $sin(100) neq 1$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      If $sin(100)=1$, then $100$ would be an integer multiple of $pi/2$.

      That is,
      $$
      frac{200}{pi}
      $$
      would have to be an integer.



      Calculating, we find $frac{200}{pi} approx 63.661... $ so it appears not to be an integer. Proving this requires bounds on $pi$.



      If we are willing to accept that
      $$
      3.14 < pi < 3.142
      $$
      then we may conclude that
      $$
      63.65372374... < frac{200}{pi} < 63.694267515...
      $$
      and since there is no integer between these two bounds, we know $frac{200}{pi}$ is not an integer, and hence $sin(100) neq 1$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        If $sin(100)=1$, then $100$ would be an integer multiple of $pi/2$.

        That is,
        $$
        frac{200}{pi}
        $$
        would have to be an integer.



        Calculating, we find $frac{200}{pi} approx 63.661... $ so it appears not to be an integer. Proving this requires bounds on $pi$.



        If we are willing to accept that
        $$
        3.14 < pi < 3.142
        $$
        then we may conclude that
        $$
        63.65372374... < frac{200}{pi} < 63.694267515...
        $$
        and since there is no integer between these two bounds, we know $frac{200}{pi}$ is not an integer, and hence $sin(100) neq 1$.






        share|cite|improve this answer









        $endgroup$



        If $sin(100)=1$, then $100$ would be an integer multiple of $pi/2$.

        That is,
        $$
        frac{200}{pi}
        $$
        would have to be an integer.



        Calculating, we find $frac{200}{pi} approx 63.661... $ so it appears not to be an integer. Proving this requires bounds on $pi$.



        If we are willing to accept that
        $$
        3.14 < pi < 3.142
        $$
        then we may conclude that
        $$
        63.65372374... < frac{200}{pi} < 63.694267515...
        $$
        and since there is no integer between these two bounds, we know $frac{200}{pi}$ is not an integer, and hence $sin(100) neq 1$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Oct 4 '16 at 2:18









        Matthew ConroyMatthew Conroy

        10.3k32836




        10.3k32836






























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