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









New contributor




airlangga is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    $a_n = (-1)^n left(frac{pi}{4}right)^n$ and $ frac{pi}{4}<1$ and this is an alternating series..... therefore by the Alternating Series Test this series.....
    $endgroup$
    – Clclstdnt
    3 hours ago












  • $begingroup$
    @Clclstdnt: this is a sequence, not a series. No alternating series test.
    $endgroup$
    – user587192
    3 hours ago












  • $begingroup$
    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
    3 hours ago












  • $begingroup$
    @user587192 I just attached her work
    $endgroup$
    – airlangga
    3 hours ago










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


















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









New contributor




airlangga is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    $a_n = (-1)^n left(frac{pi}{4}right)^n$ and $ frac{pi}{4}<1$ and this is an alternating series..... therefore by the Alternating Series Test this series.....
    $endgroup$
    – Clclstdnt
    3 hours ago












  • $begingroup$
    @Clclstdnt: this is a sequence, not a series. No alternating series test.
    $endgroup$
    – user587192
    3 hours ago












  • $begingroup$
    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
    3 hours ago












  • $begingroup$
    @user587192 I just attached her work
    $endgroup$
    – airlangga
    3 hours ago










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
















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









New contributor




airlangga is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




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









New contributor




airlangga is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




airlangga is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 2 hours ago









user587192

1,779315




1,779315






New contributor




airlangga is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 3 hours ago









airlanggaairlangga

112




112




New contributor




airlangga is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





airlangga is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






airlangga is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    $a_n = (-1)^n left(frac{pi}{4}right)^n$ and $ frac{pi}{4}<1$ and this is an alternating series..... therefore by the Alternating Series Test this series.....
    $endgroup$
    – Clclstdnt
    3 hours ago












  • $begingroup$
    @Clclstdnt: this is a sequence, not a series. No alternating series test.
    $endgroup$
    – user587192
    3 hours ago












  • $begingroup$
    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
    3 hours ago












  • $begingroup$
    @user587192 I just attached her work
    $endgroup$
    – airlangga
    3 hours ago










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




















  • $begingroup$
    $a_n = (-1)^n left(frac{pi}{4}right)^n$ and $ frac{pi}{4}<1$ and this is an alternating series..... therefore by the Alternating Series Test this series.....
    $endgroup$
    – Clclstdnt
    3 hours ago












  • $begingroup$
    @Clclstdnt: this is a sequence, not a series. No alternating series test.
    $endgroup$
    – user587192
    3 hours ago












  • $begingroup$
    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
    3 hours ago












  • $begingroup$
    @user587192 I just attached her work
    $endgroup$
    – airlangga
    3 hours ago










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


















$begingroup$
$a_n = (-1)^n left(frac{pi}{4}right)^n$ and $ frac{pi}{4}<1$ and this is an alternating series..... therefore by the Alternating Series Test this series.....
$endgroup$
– Clclstdnt
3 hours ago






$begingroup$
$a_n = (-1)^n left(frac{pi}{4}right)^n$ and $ frac{pi}{4}<1$ and this is an alternating series..... therefore by the Alternating Series Test this series.....
$endgroup$
– Clclstdnt
3 hours ago














$begingroup$
@Clclstdnt: this is a sequence, not a series. No alternating series test.
$endgroup$
– user587192
3 hours ago






$begingroup$
@Clclstdnt: this is a sequence, not a series. No alternating series test.
$endgroup$
– user587192
3 hours ago














$begingroup$
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
3 hours ago






$begingroup$
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
3 hours ago














$begingroup$
@user587192 I just attached her work
$endgroup$
– airlangga
3 hours ago




$begingroup$
@user587192 I just attached her work
$endgroup$
– airlangga
3 hours ago












$begingroup$
@airlangga: Your teacher's point is that $frac{pi}{4}<1$. And the sequence converges.
$endgroup$
– user587192
3 hours ago






$begingroup$
@airlangga: Your teacher's point is that $frac{pi}{4}<1$. And the sequence converges.
$endgroup$
– user587192
3 hours ago












2 Answers
2






active

oldest

votes


















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









$endgroup$





















    2












    $begingroup$

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






    share|cite|improve this answer









    $endgroup$













      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });






      airlangga is a new contributor. Be nice, and check out our Code of Conduct.










      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3074045%2fdetermine-whether-the-sequence-converges-or-diverges%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      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 3 hours ago









          Jimmy SabaterJimmy Sabater

          2,136219




          2,136219























              2












              $begingroup$

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






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

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






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  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 3 hours ago









                  Rhys HughesRhys Hughes

                  5,1801427




                  5,1801427






















                      airlangga is a new contributor. Be nice, and check out our Code of Conduct.










                      draft saved

                      draft discarded


















                      airlangga is a new contributor. Be nice, and check out our Code of Conduct.













                      airlangga is a new contributor. Be nice, and check out our Code of Conduct.












                      airlangga is a new contributor. Be nice, and check out our Code of Conduct.
















                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3074045%2fdetermine-whether-the-sequence-converges-or-diverges%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      Understanding the information contained in the Deep Space Network XML data?

                      Image
                      up vote 1 down vote favorite NASA's Deep Space Network activity can be viewed conveniently at https://eyes.nasa.gov/dsn/dsn.html but if you would like data in a more raw XML form, it can be accessed as described in this answer Below is a small Python script that reads the XML into a list of dictionaries, and saves it to disk as a JSON. The dictionaries are organized at the higher level by dish, then by spacecraft per dish. Question: What are the most-likely meanings of all the data field contained in the Deep Space Network XML data? This is the example url from the other answer, used in the script below: https://eyes.nasa.gov/dsn/data/1365107113.xml def dictify(r,root=True): # https://stackoverflow.com/questions/2148119/how-to-convert-an-xml-string-to-a-dictionary-in-python/30923963#30923963 if root: ret
                      Read more

                      Ross-on-Wye

                      Image
                      Ross-on-Wye Welsh: Rhosan ar Wy Town centre, looking north from Market House Ross-on-Wye Ross-on-Wye shown within Herefordshire Population 10,700 (2011) [1] OS grid reference SO597241 Unitary authority Herefordshire Ceremonial county Herefordshire Region West Midlands Country England Sovereign state United Kingdom Post town ROSS-ON-WYE Postcode district HR9 Dialling code 01989 Police West Mercia Fire Hereford and Worcester Ambulance West Midlands EU Parliament West Midlands UK Parliament Hereford and South Herefordshire List of places UK England Herefordshire 51°54′50″N 2°35′13″W  /  51.914°N 2.587°W  / 51.914; -2.587 Coordinates: 51°54′50″N 2°35′13″W  /  51.914°N 2.587°W  / 51.914; -2.587 Ross-on-Wye (Welsh: Rhosan ar Wy [2] ) is a small market town with a population of 10,700 (according to the 2011 census), in south eastern Herefordshire, England, located on the River Wye, and on
                      Read more

                      Eastern Orthodox Church

                      Image
                      Not to be confused with Oriental Orthodoxy, Eastern Christianity, or Eastern Catholic Churches. "Eastern Orthodoxy", "Orthodox Church", "Orthodox Catholic Church", and "Orthodox Christian Church" redirect here. For other uses, see Orthodox (disambiguation). This article may be too long to read and navigate comfortably . The readable prose size is 132 kilobytes. Please consider splitting content into sub-articles, condensing it, or adding or removing subheadings. ( January 2018 ) Part of a series on the Eastern Orthodox Church Mosaic of Christ Pantocrator, Hagia Sophia Overview Structure Theology (History of theology) Liturgy Church history Holy Mysteries View of salvation View of Mary View of icons Background Crucifixion / Resurrection / Ascension of Jesus Christianity Christian Church Apostolic succession Four Marks of the Church Orthodoxy Organization Autocephaly Patriarchate
                      Read more