If $sum a_k$ converges, does $sum(a_{k+1}- 2 a_{k+3})$ converge as well?












6












$begingroup$



If $sum_{k=0}^infty a_k$ is convergent with value $s$, what about $sum_{k=0}^infty b_k$ where $b_k=a_{k+1}- 2 a_{k+3}$?




My reasoning:
$$sum_{k=0}^infty b_k =lim_{n rightarrow infty} sum_{k=0}^n b_k=lim_{n rightarrow infty} sum_{k=0}^na_{k+1}-2a_{k+3}
$$
Within the sum we are only dealing with finitely many terms, we can split up the sum and take the limit afterwards:
$$ lim_{n rightarrow infty} sum_{k=0}^n b_k=lim_{n rightarrow infty} left( sum_{k=0}^n a_{k+1}- sum_{k=0}^n 2a_{k+3} right) $$
Now we want to get $s$ in here, the value of our sum, we need to do some index juggling, since our sum is not of the right form yet.
$$lim_{n rightarrow infty} left( sum_{k=0}^n a_{k} - a_0- 2sum_{k=0}^n a_{k} +2a_0 + 2a_1+2a_2right) $$
Here we applied an index shift, but then we need to compensate for the terms that we added to the sum, we finally apply the limit and get:
$$sum_{k=0}^infty b_k=s +a_0-2s+2a_1+2a_2= a_0+2a_1+2a_2 -s $$



Did that all make sense, is my reasoning correct?
conclusion: it converges as we computed the exact value.






real-analysis sequences-and-series proof-verification







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  • 2




    $begingroup$
    If $sum x_n = X in mathbb{R}$ and $sum y_n = Y in mathbb{R}$, then $sum (x_n + y_n) =X+ Y$, so yes, the reasoning is correct.
    $endgroup$
    – Cosmin
    Dec 2 '18 at 13:40








  • 1




    $begingroup$
    Looks fine to me
    $endgroup$
    – Shubham Johri
    Dec 2 '18 at 14:22












  • $begingroup$
    Please provide your answers in the answer section. This way the question does not stay open.
    $endgroup$
    – Wesley Strik
    Dec 3 '18 at 11:31










  • $begingroup$
    math.meta.stackexchange.com/questions/1559/…
    $endgroup$
    – Wesley Strik
    Dec 3 '18 at 11:33
















6












$begingroup$



If $sum_{k=0}^infty a_k$ is convergent with value $s$, what about $sum_{k=0}^infty b_k$ where $b_k=a_{k+1}- 2 a_{k+3}$?




My reasoning:
$$sum_{k=0}^infty b_k =lim_{n rightarrow infty} sum_{k=0}^n b_k=lim_{n rightarrow infty} sum_{k=0}^na_{k+1}-2a_{k+3}
$$
Within the sum we are only dealing with finitely many terms, we can split up the sum and take the limit afterwards:
$$ lim_{n rightarrow infty} sum_{k=0}^n b_k=lim_{n rightarrow infty} left( sum_{k=0}^n a_{k+1}- sum_{k=0}^n 2a_{k+3} right) $$
Now we want to get $s$ in here, the value of our sum, we need to do some index juggling, since our sum is not of the right form yet.
$$lim_{n rightarrow infty} left( sum_{k=0}^n a_{k} - a_0- 2sum_{k=0}^n a_{k} +2a_0 + 2a_1+2a_2right) $$
Here we applied an index shift, but then we need to compensate for the terms that we added to the sum, we finally apply the limit and get:
$$sum_{k=0}^infty b_k=s +a_0-2s+2a_1+2a_2= a_0+2a_1+2a_2 -s $$



Did that all make sense, is my reasoning correct?
conclusion: it converges as we computed the exact value.






real-analysis sequences-and-series proof-verification







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    If $sum x_n = X in mathbb{R}$ and $sum y_n = Y in mathbb{R}$, then $sum (x_n + y_n) =X+ Y$, so yes, the reasoning is correct.
    $endgroup$
    – Cosmin
    Dec 2 '18 at 13:40








  • 1




    $begingroup$
    Looks fine to me
    $endgroup$
    – Shubham Johri
    Dec 2 '18 at 14:22












  • $begingroup$
    Please provide your answers in the answer section. This way the question does not stay open.
    $endgroup$
    – Wesley Strik
    Dec 3 '18 at 11:31










  • $begingroup$
    math.meta.stackexchange.com/questions/1559/…
    $endgroup$
    – Wesley Strik
    Dec 3 '18 at 11:33














6












6








6





$begingroup$



If $sum_{k=0}^infty a_k$ is convergent with value $s$, what about $sum_{k=0}^infty b_k$ where $b_k=a_{k+1}- 2 a_{k+3}$?




My reasoning:
$$sum_{k=0}^infty b_k =lim_{n rightarrow infty} sum_{k=0}^n b_k=lim_{n rightarrow infty} sum_{k=0}^na_{k+1}-2a_{k+3}
$$
Within the sum we are only dealing with finitely many terms, we can split up the sum and take the limit afterwards:
$$ lim_{n rightarrow infty} sum_{k=0}^n b_k=lim_{n rightarrow infty} left( sum_{k=0}^n a_{k+1}- sum_{k=0}^n 2a_{k+3} right) $$
Now we want to get $s$ in here, the value of our sum, we need to do some index juggling, since our sum is not of the right form yet.
$$lim_{n rightarrow infty} left( sum_{k=0}^n a_{k} - a_0- 2sum_{k=0}^n a_{k} +2a_0 + 2a_1+2a_2right) $$
Here we applied an index shift, but then we need to compensate for the terms that we added to the sum, we finally apply the limit and get:
$$sum_{k=0}^infty b_k=s +a_0-2s+2a_1+2a_2= a_0+2a_1+2a_2 -s $$



Did that all make sense, is my reasoning correct?
conclusion: it converges as we computed the exact value.






real-analysis sequences-and-series proof-verification







share|cite|improve this question











$endgroup$





If $sum_{k=0}^infty a_k$ is convergent with value $s$, what about $sum_{k=0}^infty b_k$ where $b_k=a_{k+1}- 2 a_{k+3}$?




My reasoning:
$$sum_{k=0}^infty b_k =lim_{n rightarrow infty} sum_{k=0}^n b_k=lim_{n rightarrow infty} sum_{k=0}^na_{k+1}-2a_{k+3}
$$
Within the sum we are only dealing with finitely many terms, we can split up the sum and take the limit afterwards:
$$ lim_{n rightarrow infty} sum_{k=0}^n b_k=lim_{n rightarrow infty} left( sum_{k=0}^n a_{k+1}- sum_{k=0}^n 2a_{k+3} right) $$
Now we want to get $s$ in here, the value of our sum, we need to do some index juggling, since our sum is not of the right form yet.
$$lim_{n rightarrow infty} left( sum_{k=0}^n a_{k} - a_0- 2sum_{k=0}^n a_{k} +2a_0 + 2a_1+2a_2right) $$
Here we applied an index shift, but then we need to compensate for the terms that we added to the sum, we finally apply the limit and get:
$$sum_{k=0}^infty b_k=s +a_0-2s+2a_1+2a_2= a_0+2a_1+2a_2 -s $$



Did that all make sense, is my reasoning correct?
conclusion: it converges as we computed the exact value.






real-analysis sequences-and-series proof-verification




real-analysis sequences-and-series proof-verification






share|cite|improve this question















share|cite|improve this question













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edited Dec 3 '18 at 11:32









Did

247k23222458




247k23222458










asked Dec 2 '18 at 13:07









Wesley StrikWesley Strik

1,653423




1,653423








  • 2




    $begingroup$
    If $sum x_n = X in mathbb{R}$ and $sum y_n = Y in mathbb{R}$, then $sum (x_n + y_n) =X+ Y$, so yes, the reasoning is correct.
    $endgroup$
    – Cosmin
    Dec 2 '18 at 13:40








  • 1




    $begingroup$
    Looks fine to me
    $endgroup$
    – Shubham Johri
    Dec 2 '18 at 14:22












  • $begingroup$
    Please provide your answers in the answer section. This way the question does not stay open.
    $endgroup$
    – Wesley Strik
    Dec 3 '18 at 11:31










  • $begingroup$
    math.meta.stackexchange.com/questions/1559/…
    $endgroup$
    – Wesley Strik
    Dec 3 '18 at 11:33














  • 2




    $begingroup$
    If $sum x_n = X in mathbb{R}$ and $sum y_n = Y in mathbb{R}$, then $sum (x_n + y_n) =X+ Y$, so yes, the reasoning is correct.
    $endgroup$
    – Cosmin
    Dec 2 '18 at 13:40








  • 1




    $begingroup$
    Looks fine to me
    $endgroup$
    – Shubham Johri
    Dec 2 '18 at 14:22












  • $begingroup$
    Please provide your answers in the answer section. This way the question does not stay open.
    $endgroup$
    – Wesley Strik
    Dec 3 '18 at 11:31










  • $begingroup$
    math.meta.stackexchange.com/questions/1559/…
    $endgroup$
    – Wesley Strik
    Dec 3 '18 at 11:33








2




2




$begingroup$
If $sum x_n = X in mathbb{R}$ and $sum y_n = Y in mathbb{R}$, then $sum (x_n + y_n) =X+ Y$, so yes, the reasoning is correct.
$endgroup$
– Cosmin
Dec 2 '18 at 13:40






$begingroup$
If $sum x_n = X in mathbb{R}$ and $sum y_n = Y in mathbb{R}$, then $sum (x_n + y_n) =X+ Y$, so yes, the reasoning is correct.
$endgroup$
– Cosmin
Dec 2 '18 at 13:40






1




1




$begingroup$
Looks fine to me
$endgroup$
– Shubham Johri
Dec 2 '18 at 14:22






$begingroup$
Looks fine to me
$endgroup$
– Shubham Johri
Dec 2 '18 at 14:22














$begingroup$
Please provide your answers in the answer section. This way the question does not stay open.
$endgroup$
– Wesley Strik
Dec 3 '18 at 11:31




$begingroup$
Please provide your answers in the answer section. This way the question does not stay open.
$endgroup$
– Wesley Strik
Dec 3 '18 at 11:31












$begingroup$
math.meta.stackexchange.com/questions/1559/…
$endgroup$
– Wesley Strik
Dec 3 '18 at 11:33




$begingroup$
math.meta.stackexchange.com/questions/1559/…
$endgroup$
– Wesley Strik
Dec 3 '18 at 11:33










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The reasoning looks fine and correct to me.






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    I will write this in the answer section so that the question does not remain open.



    The reasoning looks fine and correct to me.






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      $begingroup$

      I will write this in the answer section so that the question does not remain open.



      The reasoning looks fine and correct to me.






      share|cite|improve this answer









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        $begingroup$

        I will write this in the answer section so that the question does not remain open.



        The reasoning looks fine and correct to me.






        share|cite|improve this answer









        $endgroup$



        I will write this in the answer section so that the question does not remain open.



        The reasoning looks fine and correct to me.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 4 '18 at 17:55









        CosminCosmin

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