Can a telescopic series with $a_n$ not tending to zero converge?

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Let $a_n$ be a sequence with $a_nto lne0$ as $ntoinfty$.



Then, does the series $$sum^infty_n=1(a_n-a_n+1)$$ necessarily converge?



I don’t know which of the following arguments is correct:



The series converges:



Because $$sum^N_n=1(a_n-a_n+1)=a_1-a_N$$
Taking limit $Ntoinfty$ on both sides, the series converges to $a_1-l$.



The series diverges:



Set $a_n=1$ for all $n$.



Then the series is equivalent to
$$1-1+1-1+1-1cdots$$ which does not converge.



Furthermore, if the series does not always converge, under what conditions would the series converge?




sequences-and-series summation




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  • 2




    Guess the sums are indexed by $n$, not by $k$. The series converges Yes, iff $a_n$ converges. The series diverges But you are not allowed to remove the parentheses, unless stricter conditions are met. Keeping the parentheses in place, the latter series is just $(1-1)+(1-1)+ldots = 0+0+ldots$
    – dxiv
    Aug 14 at 3:40







  • 3




    If $a_n=1$ for all $n$, then all of the terms in your series are $0$, not alternating $pm 1$!
    – Henning Makholm
    Aug 14 at 3:41







  • 1




    Just a small comment to illustrate the fallacy of your second argument: "one can rewrite $1/2^n = 1/2^n + 1 - 1$, so $sum_n 1/2^n = sum_n (1/2^n + 1 - 1)$ must diverge, by the same argument."
    – Clement C.
    Aug 14 at 5:47














up vote
1
down vote

favorite
1












Let $a_n$ be a sequence with $a_nto lne0$ as $ntoinfty$.



Then, does the series $$sum^infty_n=1(a_n-a_n+1)$$ necessarily converge?



I don’t know which of the following arguments is correct:



The series converges:



Because $$sum^N_n=1(a_n-a_n+1)=a_1-a_N$$
Taking limit $Ntoinfty$ on both sides, the series converges to $a_1-l$.



The series diverges:



Set $a_n=1$ for all $n$.



Then the series is equivalent to
$$1-1+1-1+1-1cdots$$ which does not converge.



Furthermore, if the series does not always converge, under what conditions would the series converge?




sequences-and-series summation




share|cite|improve this question


















  • 2




    Guess the sums are indexed by $n$, not by $k$. The series converges Yes, iff $a_n$ converges. The series diverges But you are not allowed to remove the parentheses, unless stricter conditions are met. Keeping the parentheses in place, the latter series is just $(1-1)+(1-1)+ldots = 0+0+ldots$
    – dxiv
    Aug 14 at 3:40







  • 3




    If $a_n=1$ for all $n$, then all of the terms in your series are $0$, not alternating $pm 1$!
    – Henning Makholm
    Aug 14 at 3:41







  • 1




    Just a small comment to illustrate the fallacy of your second argument: "one can rewrite $1/2^n = 1/2^n + 1 - 1$, so $sum_n 1/2^n = sum_n (1/2^n + 1 - 1)$ must diverge, by the same argument."
    – Clement C.
    Aug 14 at 5:47












up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





Let $a_n$ be a sequence with $a_nto lne0$ as $ntoinfty$.



Then, does the series $$sum^infty_n=1(a_n-a_n+1)$$ necessarily converge?



I don’t know which of the following arguments is correct:



The series converges:



Because $$sum^N_n=1(a_n-a_n+1)=a_1-a_N$$
Taking limit $Ntoinfty$ on both sides, the series converges to $a_1-l$.



The series diverges:



Set $a_n=1$ for all $n$.



Then the series is equivalent to
$$1-1+1-1+1-1cdots$$ which does not converge.



Furthermore, if the series does not always converge, under what conditions would the series converge?




sequences-and-series summation




share|cite|improve this question














Let $a_n$ be a sequence with $a_nto lne0$ as $ntoinfty$.



Then, does the series $$sum^infty_n=1(a_n-a_n+1)$$ necessarily converge?



I don’t know which of the following arguments is correct:



The series converges:



Because $$sum^N_n=1(a_n-a_n+1)=a_1-a_N$$
Taking limit $Ntoinfty$ on both sides, the series converges to $a_1-l$.



The series diverges:



Set $a_n=1$ for all $n$.



Then the series is equivalent to
$$1-1+1-1+1-1cdots$$ which does not converge.



Furthermore, if the series does not always converge, under what conditions would the series converge?





sequences-and-series summation





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edited Aug 14 at 3:55









Math Lover

12.5k21232




12.5k21232










asked Aug 14 at 3:35









Szeto

4,2411521




4,2411521







  • 2




    Guess the sums are indexed by $n$, not by $k$. The series converges Yes, iff $a_n$ converges. The series diverges But you are not allowed to remove the parentheses, unless stricter conditions are met. Keeping the parentheses in place, the latter series is just $(1-1)+(1-1)+ldots = 0+0+ldots$
    – dxiv
    Aug 14 at 3:40







  • 3




    If $a_n=1$ for all $n$, then all of the terms in your series are $0$, not alternating $pm 1$!
    – Henning Makholm
    Aug 14 at 3:41







  • 1




    Just a small comment to illustrate the fallacy of your second argument: "one can rewrite $1/2^n = 1/2^n + 1 - 1$, so $sum_n 1/2^n = sum_n (1/2^n + 1 - 1)$ must diverge, by the same argument."
    – Clement C.
    Aug 14 at 5:47












  • 2




    Guess the sums are indexed by $n$, not by $k$. The series converges Yes, iff $a_n$ converges. The series diverges But you are not allowed to remove the parentheses, unless stricter conditions are met. Keeping the parentheses in place, the latter series is just $(1-1)+(1-1)+ldots = 0+0+ldots$
    – dxiv
    Aug 14 at 3:40







  • 3




    If $a_n=1$ for all $n$, then all of the terms in your series are $0$, not alternating $pm 1$!
    – Henning Makholm
    Aug 14 at 3:41







  • 1




    Just a small comment to illustrate the fallacy of your second argument: "one can rewrite $1/2^n = 1/2^n + 1 - 1$, so $sum_n 1/2^n = sum_n (1/2^n + 1 - 1)$ must diverge, by the same argument."
    – Clement C.
    Aug 14 at 5:47







2




2




Guess the sums are indexed by $n$, not by $k$. The series converges Yes, iff $a_n$ converges. The series diverges But you are not allowed to remove the parentheses, unless stricter conditions are met. Keeping the parentheses in place, the latter series is just $(1-1)+(1-1)+ldots = 0+0+ldots$
– dxiv
Aug 14 at 3:40





Guess the sums are indexed by $n$, not by $k$. The series converges Yes, iff $a_n$ converges. The series diverges But you are not allowed to remove the parentheses, unless stricter conditions are met. Keeping the parentheses in place, the latter series is just $(1-1)+(1-1)+ldots = 0+0+ldots$
– dxiv
Aug 14 at 3:40





3




3




If $a_n=1$ for all $n$, then all of the terms in your series are $0$, not alternating $pm 1$!
– Henning Makholm
Aug 14 at 3:41





If $a_n=1$ for all $n$, then all of the terms in your series are $0$, not alternating $pm 1$!
– Henning Makholm
Aug 14 at 3:41





1




1




Just a small comment to illustrate the fallacy of your second argument: "one can rewrite $1/2^n = 1/2^n + 1 - 1$, so $sum_n 1/2^n = sum_n (1/2^n + 1 - 1)$ must diverge, by the same argument."
– Clement C.
Aug 14 at 5:47




Just a small comment to illustrate the fallacy of your second argument: "one can rewrite $1/2^n = 1/2^n + 1 - 1$, so $sum_n 1/2^n = sum_n (1/2^n + 1 - 1)$ must diverge, by the same argument."
– Clement C.
Aug 14 at 5:47










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You are assuming we don't know how the series is structured. But we do - instead of having $$1-1+1-1+cdots$$ we instead have $$(1-1)+(1-1)+cdots$$ so every term cancels and the series clearly converges! This is more evident from the sigma notation, since $a_n = a_n+1 = 1$ so $sum (a_n - a_n+1) = sum 0 = 0$






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    You must realize that the infinite series (which is rigorously denoted by a summation expression) is defined as the limit of the partial sums if it exists, and the partial sums are in turn defined in a certain precise way. One is not permitted to do any rearrangement, especially by 'intuition', unless one proves that doing so does not change the limit of the partial sums. This is why $sum_k=0^infty (-1)^k$ diverges while $sum_k=0^infty ((-1)^2k+(-1)^2k+1)$ converges.






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      You are assuming we don't know how the series is structured. But we do - instead of having $$1-1+1-1+cdots$$ we instead have $$(1-1)+(1-1)+cdots$$ so every term cancels and the series clearly converges! This is more evident from the sigma notation, since $a_n = a_n+1 = 1$ so $sum (a_n - a_n+1) = sum 0 = 0$






      share|cite|improve this answer
























        up vote
        1
        down vote













        You are assuming we don't know how the series is structured. But we do - instead of having $$1-1+1-1+cdots$$ we instead have $$(1-1)+(1-1)+cdots$$ so every term cancels and the series clearly converges! This is more evident from the sigma notation, since $a_n = a_n+1 = 1$ so $sum (a_n - a_n+1) = sum 0 = 0$






        share|cite|improve this answer






















          up vote
          1
          down vote










          up vote
          1
          down vote









          You are assuming we don't know how the series is structured. But we do - instead of having $$1-1+1-1+cdots$$ we instead have $$(1-1)+(1-1)+cdots$$ so every term cancels and the series clearly converges! This is more evident from the sigma notation, since $a_n = a_n+1 = 1$ so $sum (a_n - a_n+1) = sum 0 = 0$






          share|cite|improve this answer












          You are assuming we don't know how the series is structured. But we do - instead of having $$1-1+1-1+cdots$$ we instead have $$(1-1)+(1-1)+cdots$$ so every term cancels and the series clearly converges! This is more evident from the sigma notation, since $a_n = a_n+1 = 1$ so $sum (a_n - a_n+1) = sum 0 = 0$







          share|cite|improve this answer












          share|cite|improve this answer



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          answered Aug 14 at 5:21









          Brevan Ellefsen

          11.4k31449




          11.4k31449




















              up vote
              1
              down vote













              You must realize that the infinite series (which is rigorously denoted by a summation expression) is defined as the limit of the partial sums if it exists, and the partial sums are in turn defined in a certain precise way. One is not permitted to do any rearrangement, especially by 'intuition', unless one proves that doing so does not change the limit of the partial sums. This is why $sum_k=0^infty (-1)^k$ diverges while $sum_k=0^infty ((-1)^2k+(-1)^2k+1)$ converges.






              share|cite|improve this answer
























                up vote
                1
                down vote













                You must realize that the infinite series (which is rigorously denoted by a summation expression) is defined as the limit of the partial sums if it exists, and the partial sums are in turn defined in a certain precise way. One is not permitted to do any rearrangement, especially by 'intuition', unless one proves that doing so does not change the limit of the partial sums. This is why $sum_k=0^infty (-1)^k$ diverges while $sum_k=0^infty ((-1)^2k+(-1)^2k+1)$ converges.






                share|cite|improve this answer






















                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  You must realize that the infinite series (which is rigorously denoted by a summation expression) is defined as the limit of the partial sums if it exists, and the partial sums are in turn defined in a certain precise way. One is not permitted to do any rearrangement, especially by 'intuition', unless one proves that doing so does not change the limit of the partial sums. This is why $sum_k=0^infty (-1)^k$ diverges while $sum_k=0^infty ((-1)^2k+(-1)^2k+1)$ converges.






                  share|cite|improve this answer












                  You must realize that the infinite series (which is rigorously denoted by a summation expression) is defined as the limit of the partial sums if it exists, and the partial sums are in turn defined in a certain precise way. One is not permitted to do any rearrangement, especially by 'intuition', unless one proves that doing so does not change the limit of the partial sums. This is why $sum_k=0^infty (-1)^k$ diverges while $sum_k=0^infty ((-1)^2k+(-1)^2k+1)$ converges.







                  share|cite|improve this answer












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                  answered Aug 14 at 7:59









                  user21820

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