Sum of indicated series












0














Suppose that $sum_{n=1}^infty a_n = 1, sum_{n=1}^infty b_n = -1, a_1 = 2, b_1 = -3$



compute:
$sum_{n=1}^infty (a_n - 2b_n)$



I am a little bit stuck on what to do with all of the information given. My first idea:



$sum_{n=1}^infty (a_n) - 2sum_{n=1}^infty ( 2b_n) = 1$



I don't think that this is right because I did not use all of the information, but what would be the best way to solve this problem?






sequences-and-series







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  • 1




    the informations on $a_1$ and $b_1$ aren't usefull at all. Also $1 -2 cdot (-1) = 3 $ not $1$.
    – Thinking
    Nov 30 at 17:14








  • 1




    Doesn't the use of the most simple expedient of simply plugging the sums themselves in depend on the given two series being absolutely convergent? I think you could devise a infinitude of non -absolutely convergent series for which it would not be valid.
    – AmbretteOrrisey
    Nov 30 at 17:25


















0














Suppose that $sum_{n=1}^infty a_n = 1, sum_{n=1}^infty b_n = -1, a_1 = 2, b_1 = -3$



compute:
$sum_{n=1}^infty (a_n - 2b_n)$



I am a little bit stuck on what to do with all of the information given. My first idea:



$sum_{n=1}^infty (a_n) - 2sum_{n=1}^infty ( 2b_n) = 1$



I don't think that this is right because I did not use all of the information, but what would be the best way to solve this problem?






sequences-and-series







share|cite|improve this question


















  • 1




    the informations on $a_1$ and $b_1$ aren't usefull at all. Also $1 -2 cdot (-1) = 3 $ not $1$.
    – Thinking
    Nov 30 at 17:14








  • 1




    Doesn't the use of the most simple expedient of simply plugging the sums themselves in depend on the given two series being absolutely convergent? I think you could devise a infinitude of non -absolutely convergent series for which it would not be valid.
    – AmbretteOrrisey
    Nov 30 at 17:25
















0












0








0







Suppose that $sum_{n=1}^infty a_n = 1, sum_{n=1}^infty b_n = -1, a_1 = 2, b_1 = -3$



compute:
$sum_{n=1}^infty (a_n - 2b_n)$



I am a little bit stuck on what to do with all of the information given. My first idea:



$sum_{n=1}^infty (a_n) - 2sum_{n=1}^infty ( 2b_n) = 1$



I don't think that this is right because I did not use all of the information, but what would be the best way to solve this problem?






sequences-and-series







share|cite|improve this question













Suppose that $sum_{n=1}^infty a_n = 1, sum_{n=1}^infty b_n = -1, a_1 = 2, b_1 = -3$



compute:
$sum_{n=1}^infty (a_n - 2b_n)$



I am a little bit stuck on what to do with all of the information given. My first idea:



$sum_{n=1}^infty (a_n) - 2sum_{n=1}^infty ( 2b_n) = 1$



I don't think that this is right because I did not use all of the information, but what would be the best way to solve this problem?






sequences-and-series




sequences-and-series






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asked Nov 30 at 17:11









Elijah

626




626








  • 1




    the informations on $a_1$ and $b_1$ aren't usefull at all. Also $1 -2 cdot (-1) = 3 $ not $1$.
    – Thinking
    Nov 30 at 17:14








  • 1




    Doesn't the use of the most simple expedient of simply plugging the sums themselves in depend on the given two series being absolutely convergent? I think you could devise a infinitude of non -absolutely convergent series for which it would not be valid.
    – AmbretteOrrisey
    Nov 30 at 17:25
















  • 1




    the informations on $a_1$ and $b_1$ aren't usefull at all. Also $1 -2 cdot (-1) = 3 $ not $1$.
    – Thinking
    Nov 30 at 17:14








  • 1




    Doesn't the use of the most simple expedient of simply plugging the sums themselves in depend on the given two series being absolutely convergent? I think you could devise a infinitude of non -absolutely convergent series for which it would not be valid.
    – AmbretteOrrisey
    Nov 30 at 17:25










1




1




the informations on $a_1$ and $b_1$ aren't usefull at all. Also $1 -2 cdot (-1) = 3 $ not $1$.
– Thinking
Nov 30 at 17:14






the informations on $a_1$ and $b_1$ aren't usefull at all. Also $1 -2 cdot (-1) = 3 $ not $1$.
– Thinking
Nov 30 at 17:14






1




1




Doesn't the use of the most simple expedient of simply plugging the sums themselves in depend on the given two series being absolutely convergent? I think you could devise a infinitude of non -absolutely convergent series for which it would not be valid.
– AmbretteOrrisey
Nov 30 at 17:25






Doesn't the use of the most simple expedient of simply plugging the sums themselves in depend on the given two series being absolutely convergent? I think you could devise a infinitude of non -absolutely convergent series for which it would not be valid.
– AmbretteOrrisey
Nov 30 at 17:25












2 Answers
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We have :



$$sum_{n = 1}^infty (a_n -2b_n) = sum_{n = 1}^infty a_n - 2 cdot sum_{n = 1}^infty b_n$$



With the first two informations given we get :



$$sum_{n = 1}^infty (a_n -2b_n) = 1 -2 cdot (-1) = 3$$



Hence the informations on $a_1$ and $b_1$ aren't useful at all.






share|cite|improve this answer





























    0














    begin{align}
    sum_{n=1}^infty (a_n - 2b_n)&=sum_{n=1}^infty (a_n) - 2sum_{n=1}^infty ( b_n)\
    &=1+2=3
    end{align}






    share|cite|improve this answer





















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      2 Answers
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      active

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      2 Answers
      2






      active

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      0














      We have :



      $$sum_{n = 1}^infty (a_n -2b_n) = sum_{n = 1}^infty a_n - 2 cdot sum_{n = 1}^infty b_n$$



      With the first two informations given we get :



      $$sum_{n = 1}^infty (a_n -2b_n) = 1 -2 cdot (-1) = 3$$



      Hence the informations on $a_1$ and $b_1$ aren't useful at all.






      share|cite|improve this answer


























        0














        We have :



        $$sum_{n = 1}^infty (a_n -2b_n) = sum_{n = 1}^infty a_n - 2 cdot sum_{n = 1}^infty b_n$$



        With the first two informations given we get :



        $$sum_{n = 1}^infty (a_n -2b_n) = 1 -2 cdot (-1) = 3$$



        Hence the informations on $a_1$ and $b_1$ aren't useful at all.






        share|cite|improve this answer
























          0












          0








          0






          We have :



          $$sum_{n = 1}^infty (a_n -2b_n) = sum_{n = 1}^infty a_n - 2 cdot sum_{n = 1}^infty b_n$$



          With the first two informations given we get :



          $$sum_{n = 1}^infty (a_n -2b_n) = 1 -2 cdot (-1) = 3$$



          Hence the informations on $a_1$ and $b_1$ aren't useful at all.






          share|cite|improve this answer












          We have :



          $$sum_{n = 1}^infty (a_n -2b_n) = sum_{n = 1}^infty a_n - 2 cdot sum_{n = 1}^infty b_n$$



          With the first two informations given we get :



          $$sum_{n = 1}^infty (a_n -2b_n) = 1 -2 cdot (-1) = 3$$



          Hence the informations on $a_1$ and $b_1$ aren't useful at all.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 30 at 17:17









          Thinking

          91916




          91916























              0














              begin{align}
              sum_{n=1}^infty (a_n - 2b_n)&=sum_{n=1}^infty (a_n) - 2sum_{n=1}^infty ( b_n)\
              &=1+2=3
              end{align}






              share|cite|improve this answer


























                0














                begin{align}
                sum_{n=1}^infty (a_n - 2b_n)&=sum_{n=1}^infty (a_n) - 2sum_{n=1}^infty ( b_n)\
                &=1+2=3
                end{align}






                share|cite|improve this answer
























                  0












                  0








                  0






                  begin{align}
                  sum_{n=1}^infty (a_n - 2b_n)&=sum_{n=1}^infty (a_n) - 2sum_{n=1}^infty ( b_n)\
                  &=1+2=3
                  end{align}






                  share|cite|improve this answer












                  begin{align}
                  sum_{n=1}^infty (a_n - 2b_n)&=sum_{n=1}^infty (a_n) - 2sum_{n=1}^infty ( b_n)\
                  &=1+2=3
                  end{align}







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 30 at 17:18









                  Thomas Shelby

                  1,202116




                  1,202116






























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