Convergence and divergence of series when changing finite number of summands











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I found a text saying that adding or changing only a finite number of summands does not have an effect on the convergence/divergence of the series. This is shown by the following argumentation:



Let $sum_{k=n_1}^{infty} a_k$ and $sum_{k=n_2}^{infty}b_k$ be two series with $(s_n)_{n geq n_1}$ and $(t_n)_{t geq n_2}$ their partial sums. Let's suppose there exists an $N$ so that $a_k = b_k$ for alle $kgeq N$, than we have



begin{align}s_n = sum_{k=n_1}^{n}a_k = a_{n_1} + a_{n_1+1} + ldots + a_{N-1} + sum_{k=N}^{n}a_kend{align} and



begin{align}t_n &= sum_{k=n_2}^{n}b_k = b_{n_2} + b_{n_2+1} + ldots + b_{N-1} + sum_{k=N}^{n}a_k \
&= s_n - left(a_{n_1} + a_{n_1+1} + ldots + a_{N-1}right) + left(b_{n_2} + b_{n_2+1} + ldots + b_{N-1}right)end{align}



for all $n geq N$. Hence, both $(s_n)_{n geq n_1}$ and $(t_n)_ {tgeq n_2}$ are either convergent or divergent.



Unfortunately I do not see why $(s_n)_{n geq n_1}$ and $(t_n)_{t geq n_2}$ are either convergent or divergent following this calculation. Moreover I also don't get why this is showing that a finite number of changes to the summands of the series does not change the convergence behaviour of the series.
Can someone please help me understanding this proof.










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    Both partial series to $N-1$ contain a finite number of finite-valued terms, so these partial sums are finite. This finiteness will not affect whether the total sums (with the infinite number of terms) is finite or not.
    – Lucozade
    Nov 14 at 16:36















up vote
0
down vote

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I found a text saying that adding or changing only a finite number of summands does not have an effect on the convergence/divergence of the series. This is shown by the following argumentation:



Let $sum_{k=n_1}^{infty} a_k$ and $sum_{k=n_2}^{infty}b_k$ be two series with $(s_n)_{n geq n_1}$ and $(t_n)_{t geq n_2}$ their partial sums. Let's suppose there exists an $N$ so that $a_k = b_k$ for alle $kgeq N$, than we have



begin{align}s_n = sum_{k=n_1}^{n}a_k = a_{n_1} + a_{n_1+1} + ldots + a_{N-1} + sum_{k=N}^{n}a_kend{align} and



begin{align}t_n &= sum_{k=n_2}^{n}b_k = b_{n_2} + b_{n_2+1} + ldots + b_{N-1} + sum_{k=N}^{n}a_k \
&= s_n - left(a_{n_1} + a_{n_1+1} + ldots + a_{N-1}right) + left(b_{n_2} + b_{n_2+1} + ldots + b_{N-1}right)end{align}



for all $n geq N$. Hence, both $(s_n)_{n geq n_1}$ and $(t_n)_ {tgeq n_2}$ are either convergent or divergent.



Unfortunately I do not see why $(s_n)_{n geq n_1}$ and $(t_n)_{t geq n_2}$ are either convergent or divergent following this calculation. Moreover I also don't get why this is showing that a finite number of changes to the summands of the series does not change the convergence behaviour of the series.
Can someone please help me understanding this proof.










share|cite|improve this question




















  • 1




    Both partial series to $N-1$ contain a finite number of finite-valued terms, so these partial sums are finite. This finiteness will not affect whether the total sums (with the infinite number of terms) is finite or not.
    – Lucozade
    Nov 14 at 16:36













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I found a text saying that adding or changing only a finite number of summands does not have an effect on the convergence/divergence of the series. This is shown by the following argumentation:



Let $sum_{k=n_1}^{infty} a_k$ and $sum_{k=n_2}^{infty}b_k$ be two series with $(s_n)_{n geq n_1}$ and $(t_n)_{t geq n_2}$ their partial sums. Let's suppose there exists an $N$ so that $a_k = b_k$ for alle $kgeq N$, than we have



begin{align}s_n = sum_{k=n_1}^{n}a_k = a_{n_1} + a_{n_1+1} + ldots + a_{N-1} + sum_{k=N}^{n}a_kend{align} and



begin{align}t_n &= sum_{k=n_2}^{n}b_k = b_{n_2} + b_{n_2+1} + ldots + b_{N-1} + sum_{k=N}^{n}a_k \
&= s_n - left(a_{n_1} + a_{n_1+1} + ldots + a_{N-1}right) + left(b_{n_2} + b_{n_2+1} + ldots + b_{N-1}right)end{align}



for all $n geq N$. Hence, both $(s_n)_{n geq n_1}$ and $(t_n)_ {tgeq n_2}$ are either convergent or divergent.



Unfortunately I do not see why $(s_n)_{n geq n_1}$ and $(t_n)_{t geq n_2}$ are either convergent or divergent following this calculation. Moreover I also don't get why this is showing that a finite number of changes to the summands of the series does not change the convergence behaviour of the series.
Can someone please help me understanding this proof.










share|cite|improve this question















I found a text saying that adding or changing only a finite number of summands does not have an effect on the convergence/divergence of the series. This is shown by the following argumentation:



Let $sum_{k=n_1}^{infty} a_k$ and $sum_{k=n_2}^{infty}b_k$ be two series with $(s_n)_{n geq n_1}$ and $(t_n)_{t geq n_2}$ their partial sums. Let's suppose there exists an $N$ so that $a_k = b_k$ for alle $kgeq N$, than we have



begin{align}s_n = sum_{k=n_1}^{n}a_k = a_{n_1} + a_{n_1+1} + ldots + a_{N-1} + sum_{k=N}^{n}a_kend{align} and



begin{align}t_n &= sum_{k=n_2}^{n}b_k = b_{n_2} + b_{n_2+1} + ldots + b_{N-1} + sum_{k=N}^{n}a_k \
&= s_n - left(a_{n_1} + a_{n_1+1} + ldots + a_{N-1}right) + left(b_{n_2} + b_{n_2+1} + ldots + b_{N-1}right)end{align}



for all $n geq N$. Hence, both $(s_n)_{n geq n_1}$ and $(t_n)_ {tgeq n_2}$ are either convergent or divergent.



Unfortunately I do not see why $(s_n)_{n geq n_1}$ and $(t_n)_{t geq n_2}$ are either convergent or divergent following this calculation. Moreover I also don't get why this is showing that a finite number of changes to the summands of the series does not change the convergence behaviour of the series.
Can someone please help me understanding this proof.







real-analysis sequences-and-series convergence






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edited Nov 14 at 16:24

























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  • 1




    Both partial series to $N-1$ contain a finite number of finite-valued terms, so these partial sums are finite. This finiteness will not affect whether the total sums (with the infinite number of terms) is finite or not.
    – Lucozade
    Nov 14 at 16:36














  • 1




    Both partial series to $N-1$ contain a finite number of finite-valued terms, so these partial sums are finite. This finiteness will not affect whether the total sums (with the infinite number of terms) is finite or not.
    – Lucozade
    Nov 14 at 16:36








1




1




Both partial series to $N-1$ contain a finite number of finite-valued terms, so these partial sums are finite. This finiteness will not affect whether the total sums (with the infinite number of terms) is finite or not.
– Lucozade
Nov 14 at 16:36




Both partial series to $N-1$ contain a finite number of finite-valued terms, so these partial sums are finite. This finiteness will not affect whether the total sums (with the infinite number of terms) is finite or not.
– Lucozade
Nov 14 at 16:36










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Suppose $lim_{n to infty}t_nto t$,
and
begin{align}t_n
&= s_n - left(a_{n_1} + a_{n_1+1} + ldots + a_{N-1}right) + left(b_{n_2} + b_{n_2+1} + ldots + b_{N-1}right).end{align}



We can let $left(a_{n_1} + a_{n_1+1} + ldots + a_{N-1}right) - left(b_{n_2} + b_{n_2+1} + ldots + b_{N-1}right)=C,$ a constant that is independent of $n$, then we have



$$t_n = s_n - C$$



then we have



begin{align}lim_{n to infty}t_n
&= lim_{n to infty}s_n - Cend{align}



Hence begin{align}
lim_{n to infty}s_n =t+ Cend{align}



That is $s_n$ converges as well.



Similarly, we can argue that if $s_n$ converges, then $t_n$ converges.






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    Suppose $lim_{n to infty}t_nto t$,
    and
    begin{align}t_n
    &= s_n - left(a_{n_1} + a_{n_1+1} + ldots + a_{N-1}right) + left(b_{n_2} + b_{n_2+1} + ldots + b_{N-1}right).end{align}



    We can let $left(a_{n_1} + a_{n_1+1} + ldots + a_{N-1}right) - left(b_{n_2} + b_{n_2+1} + ldots + b_{N-1}right)=C,$ a constant that is independent of $n$, then we have



    $$t_n = s_n - C$$



    then we have



    begin{align}lim_{n to infty}t_n
    &= lim_{n to infty}s_n - Cend{align}



    Hence begin{align}
    lim_{n to infty}s_n =t+ Cend{align}



    That is $s_n$ converges as well.



    Similarly, we can argue that if $s_n$ converges, then $t_n$ converges.






    share|cite|improve this answer

























      up vote
      1
      down vote



      accepted










      Suppose $lim_{n to infty}t_nto t$,
      and
      begin{align}t_n
      &= s_n - left(a_{n_1} + a_{n_1+1} + ldots + a_{N-1}right) + left(b_{n_2} + b_{n_2+1} + ldots + b_{N-1}right).end{align}



      We can let $left(a_{n_1} + a_{n_1+1} + ldots + a_{N-1}right) - left(b_{n_2} + b_{n_2+1} + ldots + b_{N-1}right)=C,$ a constant that is independent of $n$, then we have



      $$t_n = s_n - C$$



      then we have



      begin{align}lim_{n to infty}t_n
      &= lim_{n to infty}s_n - Cend{align}



      Hence begin{align}
      lim_{n to infty}s_n =t+ Cend{align}



      That is $s_n$ converges as well.



      Similarly, we can argue that if $s_n$ converges, then $t_n$ converges.






      share|cite|improve this answer























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        Suppose $lim_{n to infty}t_nto t$,
        and
        begin{align}t_n
        &= s_n - left(a_{n_1} + a_{n_1+1} + ldots + a_{N-1}right) + left(b_{n_2} + b_{n_2+1} + ldots + b_{N-1}right).end{align}



        We can let $left(a_{n_1} + a_{n_1+1} + ldots + a_{N-1}right) - left(b_{n_2} + b_{n_2+1} + ldots + b_{N-1}right)=C,$ a constant that is independent of $n$, then we have



        $$t_n = s_n - C$$



        then we have



        begin{align}lim_{n to infty}t_n
        &= lim_{n to infty}s_n - Cend{align}



        Hence begin{align}
        lim_{n to infty}s_n =t+ Cend{align}



        That is $s_n$ converges as well.



        Similarly, we can argue that if $s_n$ converges, then $t_n$ converges.






        share|cite|improve this answer












        Suppose $lim_{n to infty}t_nto t$,
        and
        begin{align}t_n
        &= s_n - left(a_{n_1} + a_{n_1+1} + ldots + a_{N-1}right) + left(b_{n_2} + b_{n_2+1} + ldots + b_{N-1}right).end{align}



        We can let $left(a_{n_1} + a_{n_1+1} + ldots + a_{N-1}right) - left(b_{n_2} + b_{n_2+1} + ldots + b_{N-1}right)=C,$ a constant that is independent of $n$, then we have



        $$t_n = s_n - C$$



        then we have



        begin{align}lim_{n to infty}t_n
        &= lim_{n to infty}s_n - Cend{align}



        Hence begin{align}
        lim_{n to infty}s_n =t+ Cend{align}



        That is $s_n$ converges as well.



        Similarly, we can argue that if $s_n$ converges, then $t_n$ converges.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 14 at 16:44









        Siong Thye Goh

        93.6k1462114




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