Does $sum _{n=1}^{infty ::}left(frac{-3}{4}right)^{n+1}$ converge?












2














This is a geometric series, where $a=(3/4)^2$ and $r=-3/4$, meaning that it converges to $9/28$ right? The answer in my textbook says that it diverges, whereas Symbolab says it converges absolutely. So does it converge or diverge?






sequences-and-series limits







share|cite|improve this question




















  • 4




    It most definitely converges since it's a geometric series whose ratio has absolute value less than $1$.
    – Fimpellizieri
    Nov 23 at 3:26










  • Absolute convergence is a very strong result in analysis. So basically, IF you can show that $sum_{n=1}^infty |a_n|$ converges, then you know the sum without absolute value $sum_{n=1}^infty a_n$ WILL converge ALSO. It makes showing convergence of some sums much easier.
    – Decaf-Math
    Nov 23 at 3:46










  • Your textbook probably has a typo.
    – Paramanand Singh
    Nov 23 at 12:25
















2














This is a geometric series, where $a=(3/4)^2$ and $r=-3/4$, meaning that it converges to $9/28$ right? The answer in my textbook says that it diverges, whereas Symbolab says it converges absolutely. So does it converge or diverge?






sequences-and-series limits







share|cite|improve this question




















  • 4




    It most definitely converges since it's a geometric series whose ratio has absolute value less than $1$.
    – Fimpellizieri
    Nov 23 at 3:26










  • Absolute convergence is a very strong result in analysis. So basically, IF you can show that $sum_{n=1}^infty |a_n|$ converges, then you know the sum without absolute value $sum_{n=1}^infty a_n$ WILL converge ALSO. It makes showing convergence of some sums much easier.
    – Decaf-Math
    Nov 23 at 3:46










  • Your textbook probably has a typo.
    – Paramanand Singh
    Nov 23 at 12:25














2












2








2







This is a geometric series, where $a=(3/4)^2$ and $r=-3/4$, meaning that it converges to $9/28$ right? The answer in my textbook says that it diverges, whereas Symbolab says it converges absolutely. So does it converge or diverge?






sequences-and-series limits







share|cite|improve this question















This is a geometric series, where $a=(3/4)^2$ and $r=-3/4$, meaning that it converges to $9/28$ right? The answer in my textbook says that it diverges, whereas Symbolab says it converges absolutely. So does it converge or diverge?






sequences-and-series limits




sequences-and-series limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 23 at 3:28









Seth

42812




42812










asked Nov 23 at 3:23









Fourth

514




514








  • 4




    It most definitely converges since it's a geometric series whose ratio has absolute value less than $1$.
    – Fimpellizieri
    Nov 23 at 3:26










  • Absolute convergence is a very strong result in analysis. So basically, IF you can show that $sum_{n=1}^infty |a_n|$ converges, then you know the sum without absolute value $sum_{n=1}^infty a_n$ WILL converge ALSO. It makes showing convergence of some sums much easier.
    – Decaf-Math
    Nov 23 at 3:46










  • Your textbook probably has a typo.
    – Paramanand Singh
    Nov 23 at 12:25














  • 4




    It most definitely converges since it's a geometric series whose ratio has absolute value less than $1$.
    – Fimpellizieri
    Nov 23 at 3:26










  • Absolute convergence is a very strong result in analysis. So basically, IF you can show that $sum_{n=1}^infty |a_n|$ converges, then you know the sum without absolute value $sum_{n=1}^infty a_n$ WILL converge ALSO. It makes showing convergence of some sums much easier.
    – Decaf-Math
    Nov 23 at 3:46










  • Your textbook probably has a typo.
    – Paramanand Singh
    Nov 23 at 12:25








4




4




It most definitely converges since it's a geometric series whose ratio has absolute value less than $1$.
– Fimpellizieri
Nov 23 at 3:26




It most definitely converges since it's a geometric series whose ratio has absolute value less than $1$.
– Fimpellizieri
Nov 23 at 3:26












Absolute convergence is a very strong result in analysis. So basically, IF you can show that $sum_{n=1}^infty |a_n|$ converges, then you know the sum without absolute value $sum_{n=1}^infty a_n$ WILL converge ALSO. It makes showing convergence of some sums much easier.
– Decaf-Math
Nov 23 at 3:46




Absolute convergence is a very strong result in analysis. So basically, IF you can show that $sum_{n=1}^infty |a_n|$ converges, then you know the sum without absolute value $sum_{n=1}^infty a_n$ WILL converge ALSO. It makes showing convergence of some sums much easier.
– Decaf-Math
Nov 23 at 3:46












Your textbook probably has a typo.
– Paramanand Singh
Nov 23 at 12:25




Your textbook probably has a typo.
– Paramanand Singh
Nov 23 at 12:25










4 Answers
4






active

oldest

votes


















2














This is an absolutely convergent series, therefore the series has to converge regardless of the minus signs.






share|cite|improve this answer





























    1














    If $|x| < 1$ then
    $displaystylesum_{n=0}^{infty} x^n
    =frac1{1-x}$    .



    Therefore



    $begin{align}\
    sum_{n=1}^{infty} left(frac{-3}{4}right)^{n+1}
    &=left(frac{-3}{4}right)^2sum_{n=1}^{infty} left(frac{-3}{4}right)^{n-1}\
    &=frac{9}{16}sum_{n=0}^{infty} left(frac{-3}{4}right)^{n}\
    &=frac{9}{16}frac1{1-left(frac{-3}{4}right)}\
    &=frac{9}{16}frac{4}{4+3}\
    &=frac{9}{4cdot 7}\
    &=frac{9}{28}\
    end{align}
    $






    share|cite|improve this answer































      1














      The series can be written as the sum of positive terms and negative terms.



      $$sum_{i=1}^{infty}(-3/4)^{i+1} = sum_{k=0}^{infty}(3/4)^{2k}-sum_{k=0}^{infty}(3/4)^{2k+1} = a_n - b_n$$



      Both $a_n$ and $b_n$ being geometric series are convergent.



      Hence, their sum should also be a convergent series. Proof of convergence of sum of two convergent series.






      share|cite|improve this answer































        1














        The Leibniz criterion could be mentioned.



        The series is convergent.



        https://en.m.wikipedia.org/wiki/Alternating_series_test






        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














          This is an absolutely convergent series, therefore the series has to converge regardless of the minus signs.






          share|cite|improve this answer


























            2














            This is an absolutely convergent series, therefore the series has to converge regardless of the minus signs.






            share|cite|improve this answer
























              2












              2








              2






              This is an absolutely convergent series, therefore the series has to converge regardless of the minus signs.






              share|cite|improve this answer












              This is an absolutely convergent series, therefore the series has to converge regardless of the minus signs.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 23 at 3:29









              Juan123

              815




              815























                  1














                  If $|x| < 1$ then
                  $displaystylesum_{n=0}^{infty} x^n
                  =frac1{1-x}$                  .



                  Therefore



                  $begin{align}\
                  sum_{n=1}^{infty} left(frac{-3}{4}right)^{n+1}
                  &=left(frac{-3}{4}right)^2sum_{n=1}^{infty} left(frac{-3}{4}right)^{n-1}\
                  &=frac{9}{16}sum_{n=0}^{infty} left(frac{-3}{4}right)^{n}\
                  &=frac{9}{16}frac1{1-left(frac{-3}{4}right)}\
                  &=frac{9}{16}frac{4}{4+3}\
                  &=frac{9}{4cdot 7}\
                  &=frac{9}{28}\
                  end{align}
                  $






                  share|cite|improve this answer




























                    1














                    If $|x| < 1$ then
                    $displaystylesum_{n=0}^{infty} x^n
                    =frac1{1-x}$                    .



                    Therefore



                    $begin{align}\
                    sum_{n=1}^{infty} left(frac{-3}{4}right)^{n+1}
                    &=left(frac{-3}{4}right)^2sum_{n=1}^{infty} left(frac{-3}{4}right)^{n-1}\
                    &=frac{9}{16}sum_{n=0}^{infty} left(frac{-3}{4}right)^{n}\
                    &=frac{9}{16}frac1{1-left(frac{-3}{4}right)}\
                    &=frac{9}{16}frac{4}{4+3}\
                    &=frac{9}{4cdot 7}\
                    &=frac{9}{28}\
                    end{align}
                    $






                    share|cite|improve this answer


























                      1












                      1








                      1






                      If $|x| < 1$ then
                      $displaystylesum_{n=0}^{infty} x^n
                      =frac1{1-x}$                      .



                      Therefore



                      $begin{align}\
                      sum_{n=1}^{infty} left(frac{-3}{4}right)^{n+1}
                      &=left(frac{-3}{4}right)^2sum_{n=1}^{infty} left(frac{-3}{4}right)^{n-1}\
                      &=frac{9}{16}sum_{n=0}^{infty} left(frac{-3}{4}right)^{n}\
                      &=frac{9}{16}frac1{1-left(frac{-3}{4}right)}\
                      &=frac{9}{16}frac{4}{4+3}\
                      &=frac{9}{4cdot 7}\
                      &=frac{9}{28}\
                      end{align}
                      $






                      share|cite|improve this answer














                      If $|x| < 1$ then
                      $displaystylesum_{n=0}^{infty} x^n
                      =frac1{1-x}$                      .



                      Therefore



                      $begin{align}\
                      sum_{n=1}^{infty} left(frac{-3}{4}right)^{n+1}
                      &=left(frac{-3}{4}right)^2sum_{n=1}^{infty} left(frac{-3}{4}right)^{n-1}\
                      &=frac{9}{16}sum_{n=0}^{infty} left(frac{-3}{4}right)^{n}\
                      &=frac{9}{16}frac1{1-left(frac{-3}{4}right)}\
                      &=frac{9}{16}frac{4}{4+3}\
                      &=frac{9}{4cdot 7}\
                      &=frac{9}{28}\
                      end{align}
                      $







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Nov 23 at 4:16









                      Chickenmancer

                      3,309723




                      3,309723










                      answered Nov 23 at 4:14









                      marty cohen

                      72.3k549127




                      72.3k549127























                          1














                          The series can be written as the sum of positive terms and negative terms.



                          $$sum_{i=1}^{infty}(-3/4)^{i+1} = sum_{k=0}^{infty}(3/4)^{2k}-sum_{k=0}^{infty}(3/4)^{2k+1} = a_n - b_n$$



                          Both $a_n$ and $b_n$ being geometric series are convergent.



                          Hence, their sum should also be a convergent series. Proof of convergence of sum of two convergent series.






                          share|cite|improve this answer




























                            1














                            The series can be written as the sum of positive terms and negative terms.



                            $$sum_{i=1}^{infty}(-3/4)^{i+1} = sum_{k=0}^{infty}(3/4)^{2k}-sum_{k=0}^{infty}(3/4)^{2k+1} = a_n - b_n$$



                            Both $a_n$ and $b_n$ being geometric series are convergent.



                            Hence, their sum should also be a convergent series. Proof of convergence of sum of two convergent series.






                            share|cite|improve this answer


























                              1












                              1








                              1






                              The series can be written as the sum of positive terms and negative terms.



                              $$sum_{i=1}^{infty}(-3/4)^{i+1} = sum_{k=0}^{infty}(3/4)^{2k}-sum_{k=0}^{infty}(3/4)^{2k+1} = a_n - b_n$$



                              Both $a_n$ and $b_n$ being geometric series are convergent.



                              Hence, their sum should also be a convergent series. Proof of convergence of sum of two convergent series.






                              share|cite|improve this answer














                              The series can be written as the sum of positive terms and negative terms.



                              $$sum_{i=1}^{infty}(-3/4)^{i+1} = sum_{k=0}^{infty}(3/4)^{2k}-sum_{k=0}^{infty}(3/4)^{2k+1} = a_n - b_n$$



                              Both $a_n$ and $b_n$ being geometric series are convergent.



                              Hence, their sum should also be a convergent series. Proof of convergence of sum of two convergent series.







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited Nov 23 at 4:16









                              Chickenmancer

                              3,309723




                              3,309723










                              answered Nov 23 at 3:52









                              Ajit Sharma

                              213




                              213























                                  1














                                  The Leibniz criterion could be mentioned.



                                  The series is convergent.



                                  https://en.m.wikipedia.org/wiki/Alternating_series_test






                                  share|cite|improve this answer


























                                    1














                                    The Leibniz criterion could be mentioned.



                                    The series is convergent.



                                    https://en.m.wikipedia.org/wiki/Alternating_series_test






                                    share|cite|improve this answer
























                                      1












                                      1








                                      1






                                      The Leibniz criterion could be mentioned.



                                      The series is convergent.



                                      https://en.m.wikipedia.org/wiki/Alternating_series_test






                                      share|cite|improve this answer












                                      The Leibniz criterion could be mentioned.



                                      The series is convergent.



                                      https://en.m.wikipedia.org/wiki/Alternating_series_test







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Nov 23 at 9:01









                                      Peter Szilas

                                      10.6k2720




                                      10.6k2720






























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