Determine whether the sequence converges or diverges.












2












$begingroup$


Let : $\a_n=frac{(-pi)^n}{4^n}$



What my teacher told me :



$a_1=frac{-pi}{4} approx -0.785\
a_2=frac{pi^2}{16} approx 0.617\
a_3=frac{-pi^3}{64} approx -0.484\
a_4=frac{pi^4}{256} approx 0.380\$



So the sequence diverges. But I'm not really sure about the answer.



Here is my teacher's work:



Here's my teacher's work






calculus sequences-and-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    @Clclstdnt: this is a sequence, not a series. No alternating series test.
    $endgroup$
    – user587192
    Jan 15 at 3:50








  • 1




    $begingroup$
    One would not simply list four terms and claim the sequence is divergent or convergent. You may have probably taken a wrong note on what your teacher said.
    $endgroup$
    – user587192
    Jan 15 at 3:52












  • $begingroup$
    @user587192 I just attached her work
    $endgroup$
    – airlangga
    Jan 15 at 4:22










  • $begingroup$
    @airlangga: Your teacher's point is that $frac{pi}{4}<1$. And the sequence converges.
    $endgroup$
    – user587192
    Jan 15 at 4:26


















2












$begingroup$


Let : $\a_n=frac{(-pi)^n}{4^n}$



What my teacher told me :



$a_1=frac{-pi}{4} approx -0.785\
a_2=frac{pi^2}{16} approx 0.617\
a_3=frac{-pi^3}{64} approx -0.484\
a_4=frac{pi^4}{256} approx 0.380\$



So the sequence diverges. But I'm not really sure about the answer.



Here is my teacher's work:



Here's my teacher's work






calculus sequences-and-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    @Clclstdnt: this is a sequence, not a series. No alternating series test.
    $endgroup$
    – user587192
    Jan 15 at 3:50








  • 1




    $begingroup$
    One would not simply list four terms and claim the sequence is divergent or convergent. You may have probably taken a wrong note on what your teacher said.
    $endgroup$
    – user587192
    Jan 15 at 3:52












  • $begingroup$
    @user587192 I just attached her work
    $endgroup$
    – airlangga
    Jan 15 at 4:22










  • $begingroup$
    @airlangga: Your teacher's point is that $frac{pi}{4}<1$. And the sequence converges.
    $endgroup$
    – user587192
    Jan 15 at 4:26
















2












2








2





$begingroup$


Let : $\a_n=frac{(-pi)^n}{4^n}$



What my teacher told me :



$a_1=frac{-pi}{4} approx -0.785\
a_2=frac{pi^2}{16} approx 0.617\
a_3=frac{-pi^3}{64} approx -0.484\
a_4=frac{pi^4}{256} approx 0.380\$



So the sequence diverges. But I'm not really sure about the answer.



Here is my teacher's work:



Here's my teacher's work






calculus sequences-and-series







share|cite|improve this question











$endgroup$




Let : $\a_n=frac{(-pi)^n}{4^n}$



What my teacher told me :



$a_1=frac{-pi}{4} approx -0.785\
a_2=frac{pi^2}{16} approx 0.617\
a_3=frac{-pi^3}{64} approx -0.484\
a_4=frac{pi^4}{256} approx 0.380\$



So the sequence diverges. But I'm not really sure about the answer.



Here is my teacher's work:



Here's my teacher's work






calculus sequences-and-series




calculus sequences-and-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 15 at 5:06









user587192

2,064315




2,064315










asked Jan 15 at 3:42









airlanggaairlangga

604




604












  • $begingroup$
    @Clclstdnt: this is a sequence, not a series. No alternating series test.
    $endgroup$
    – user587192
    Jan 15 at 3:50








  • 1




    $begingroup$
    One would not simply list four terms and claim the sequence is divergent or convergent. You may have probably taken a wrong note on what your teacher said.
    $endgroup$
    – user587192
    Jan 15 at 3:52












  • $begingroup$
    @user587192 I just attached her work
    $endgroup$
    – airlangga
    Jan 15 at 4:22










  • $begingroup$
    @airlangga: Your teacher's point is that $frac{pi}{4}<1$. And the sequence converges.
    $endgroup$
    – user587192
    Jan 15 at 4:26




















  • $begingroup$
    @Clclstdnt: this is a sequence, not a series. No alternating series test.
    $endgroup$
    – user587192
    Jan 15 at 3:50








  • 1




    $begingroup$
    One would not simply list four terms and claim the sequence is divergent or convergent. You may have probably taken a wrong note on what your teacher said.
    $endgroup$
    – user587192
    Jan 15 at 3:52












  • $begingroup$
    @user587192 I just attached her work
    $endgroup$
    – airlangga
    Jan 15 at 4:22










  • $begingroup$
    @airlangga: Your teacher's point is that $frac{pi}{4}<1$. And the sequence converges.
    $endgroup$
    – user587192
    Jan 15 at 4:26


















$begingroup$
@Clclstdnt: this is a sequence, not a series. No alternating series test.
$endgroup$
– user587192
Jan 15 at 3:50






$begingroup$
@Clclstdnt: this is a sequence, not a series. No alternating series test.
$endgroup$
– user587192
Jan 15 at 3:50






1




1




$begingroup$
One would not simply list four terms and claim the sequence is divergent or convergent. You may have probably taken a wrong note on what your teacher said.
$endgroup$
– user587192
Jan 15 at 3:52






$begingroup$
One would not simply list four terms and claim the sequence is divergent or convergent. You may have probably taken a wrong note on what your teacher said.
$endgroup$
– user587192
Jan 15 at 3:52














$begingroup$
@user587192 I just attached her work
$endgroup$
– airlangga
Jan 15 at 4:22




$begingroup$
@user587192 I just attached her work
$endgroup$
– airlangga
Jan 15 at 4:22












$begingroup$
@airlangga: Your teacher's point is that $frac{pi}{4}<1$. And the sequence converges.
$endgroup$
– user587192
Jan 15 at 4:26






$begingroup$
@airlangga: Your teacher's point is that $frac{pi}{4}<1$. And the sequence converges.
$endgroup$
– user587192
Jan 15 at 4:26












2 Answers
2






active

oldest

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3












$begingroup$

By triangle inequality one has that if $lim_{n to infty} |a_n| = 0$, then $lim_{n to infty} a_n = 0$. Thus, if $a_n = frac{(- pi )^n }{4^n } = frac{ (-1)^n pi^n }{4^n} $. Observe that



$$ left| frac{ (-1)^n pi^n }{4^n} right| = frac{ pi^n }{4^n} = left( frac{ pi }{4} right)^n to 0 $$






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    Well: $$|t|<1to lim_{ntoinfty}t^n=0$$






    share|cite|improve this answer









    $endgroup$













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      2 Answers
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      2 Answers
      2






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      active

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      active

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      3












      $begingroup$

      By triangle inequality one has that if $lim_{n to infty} |a_n| = 0$, then $lim_{n to infty} a_n = 0$. Thus, if $a_n = frac{(- pi )^n }{4^n } = frac{ (-1)^n pi^n }{4^n} $. Observe that



      $$ left| frac{ (-1)^n pi^n }{4^n} right| = frac{ pi^n }{4^n} = left( frac{ pi }{4} right)^n to 0 $$






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        By triangle inequality one has that if $lim_{n to infty} |a_n| = 0$, then $lim_{n to infty} a_n = 0$. Thus, if $a_n = frac{(- pi )^n }{4^n } = frac{ (-1)^n pi^n }{4^n} $. Observe that



        $$ left| frac{ (-1)^n pi^n }{4^n} right| = frac{ pi^n }{4^n} = left( frac{ pi }{4} right)^n to 0 $$






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          By triangle inequality one has that if $lim_{n to infty} |a_n| = 0$, then $lim_{n to infty} a_n = 0$. Thus, if $a_n = frac{(- pi )^n }{4^n } = frac{ (-1)^n pi^n }{4^n} $. Observe that



          $$ left| frac{ (-1)^n pi^n }{4^n} right| = frac{ pi^n }{4^n} = left( frac{ pi }{4} right)^n to 0 $$






          share|cite|improve this answer









          $endgroup$



          By triangle inequality one has that if $lim_{n to infty} |a_n| = 0$, then $lim_{n to infty} a_n = 0$. Thus, if $a_n = frac{(- pi )^n }{4^n } = frac{ (-1)^n pi^n }{4^n} $. Observe that



          $$ left| frac{ (-1)^n pi^n }{4^n} right| = frac{ pi^n }{4^n} = left( frac{ pi }{4} right)^n to 0 $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 15 at 3:48









          Jimmy SabaterJimmy Sabater

          2,631321




          2,631321























              3












              $begingroup$

              Well: $$|t|<1to lim_{ntoinfty}t^n=0$$






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                Well: $$|t|<1to lim_{ntoinfty}t^n=0$$






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  Well: $$|t|<1to lim_{ntoinfty}t^n=0$$






                  share|cite|improve this answer









                  $endgroup$



                  Well: $$|t|<1to lim_{ntoinfty}t^n=0$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 15 at 4:04









                  Rhys HughesRhys Hughes

                  5,9981529




                  5,9981529






























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