Convergence test for partial sum whose elements all change as the index increases

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I have a sequence of length $n$, $b_i(n),, i=1,...,n$. Each $b_i(n)$ is a strictly positive rational number. I write $b_i(n)$ because indeed each element of the sequence is a function also of the length of the sequence, and as the sequence progresses all its elements change value.



If $S_n = sum_i=1^nb_i(n)$, I need to examine whether $lim_nto inftyS_n$ converges to a finite limit or not. Partial sums look like



$$S_n = sum_i=1^nb_i(n)$$
$$S_n+1 = sum_i=1^nb_i(n+1) + b_n+1(n+1)$$



I know that the limit of the partial sum will never be zero. I know that each element of the sequence tends to zero as $n$ progresses. But since I have infinite elements of the partial sum that go to zero, I am not sure how to determine whether the whole infinite series remains finite or not.



Can somebody suggest any method or point to relevant sources that examine this issue? I am not sure that the usual tests for convergence of a series/limit of a partial sum (like the ratio test etc) apply here.







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  • 1




    I just want to make sure I understand. I'm think of this as an infinite lower triangular matrix, where $S_n$ is the sum of the $n$th row. When you say that each element of the sequence goes to $0,$ it means that each of the columns goes to $0$. Then you want to know whether that guarantees that the row sums go to a finite limit. Am I correct?
    – saulspatz
    Aug 2 at 2:17











  • This cannot be answered in general without more specifics or, even better, some example(s). each element of the sequence is a function also of the length of the sequence This is always the case for Riemann sums for example, see e.g. How do you calculate this limit $lim_ntoinftysum_k=1^n frackn^2+k^2$?, but this is by no means a general answer.
    – dxiv
    Aug 2 at 2:42











  • @saulspatz Yes, good image.
    – Alecos Papadopoulos
    Aug 2 at 5:34














up vote
0
down vote

favorite












I have a sequence of length $n$, $b_i(n),, i=1,...,n$. Each $b_i(n)$ is a strictly positive rational number. I write $b_i(n)$ because indeed each element of the sequence is a function also of the length of the sequence, and as the sequence progresses all its elements change value.



If $S_n = sum_i=1^nb_i(n)$, I need to examine whether $lim_nto inftyS_n$ converges to a finite limit or not. Partial sums look like



$$S_n = sum_i=1^nb_i(n)$$
$$S_n+1 = sum_i=1^nb_i(n+1) + b_n+1(n+1)$$



I know that the limit of the partial sum will never be zero. I know that each element of the sequence tends to zero as $n$ progresses. But since I have infinite elements of the partial sum that go to zero, I am not sure how to determine whether the whole infinite series remains finite or not.



Can somebody suggest any method or point to relevant sources that examine this issue? I am not sure that the usual tests for convergence of a series/limit of a partial sum (like the ratio test etc) apply here.







share|cite|improve this question

















  • 1




    I just want to make sure I understand. I'm think of this as an infinite lower triangular matrix, where $S_n$ is the sum of the $n$th row. When you say that each element of the sequence goes to $0,$ it means that each of the columns goes to $0$. Then you want to know whether that guarantees that the row sums go to a finite limit. Am I correct?
    – saulspatz
    Aug 2 at 2:17











  • This cannot be answered in general without more specifics or, even better, some example(s). each element of the sequence is a function also of the length of the sequence This is always the case for Riemann sums for example, see e.g. How do you calculate this limit $lim_ntoinftysum_k=1^n frackn^2+k^2$?, but this is by no means a general answer.
    – dxiv
    Aug 2 at 2:42











  • @saulspatz Yes, good image.
    – Alecos Papadopoulos
    Aug 2 at 5:34












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I have a sequence of length $n$, $b_i(n),, i=1,...,n$. Each $b_i(n)$ is a strictly positive rational number. I write $b_i(n)$ because indeed each element of the sequence is a function also of the length of the sequence, and as the sequence progresses all its elements change value.



If $S_n = sum_i=1^nb_i(n)$, I need to examine whether $lim_nto inftyS_n$ converges to a finite limit or not. Partial sums look like



$$S_n = sum_i=1^nb_i(n)$$
$$S_n+1 = sum_i=1^nb_i(n+1) + b_n+1(n+1)$$



I know that the limit of the partial sum will never be zero. I know that each element of the sequence tends to zero as $n$ progresses. But since I have infinite elements of the partial sum that go to zero, I am not sure how to determine whether the whole infinite series remains finite or not.



Can somebody suggest any method or point to relevant sources that examine this issue? I am not sure that the usual tests for convergence of a series/limit of a partial sum (like the ratio test etc) apply here.







share|cite|improve this question













I have a sequence of length $n$, $b_i(n),, i=1,...,n$. Each $b_i(n)$ is a strictly positive rational number. I write $b_i(n)$ because indeed each element of the sequence is a function also of the length of the sequence, and as the sequence progresses all its elements change value.



If $S_n = sum_i=1^nb_i(n)$, I need to examine whether $lim_nto inftyS_n$ converges to a finite limit or not. Partial sums look like



$$S_n = sum_i=1^nb_i(n)$$
$$S_n+1 = sum_i=1^nb_i(n+1) + b_n+1(n+1)$$



I know that the limit of the partial sum will never be zero. I know that each element of the sequence tends to zero as $n$ progresses. But since I have infinite elements of the partial sum that go to zero, I am not sure how to determine whether the whole infinite series remains finite or not.



Can somebody suggest any method or point to relevant sources that examine this issue? I am not sure that the usual tests for convergence of a series/limit of a partial sum (like the ratio test etc) apply here.









share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Aug 2 at 5:36
























asked Aug 2 at 0:17









Alecos Papadopoulos

7,92811535




7,92811535







  • 1




    I just want to make sure I understand. I'm think of this as an infinite lower triangular matrix, where $S_n$ is the sum of the $n$th row. When you say that each element of the sequence goes to $0,$ it means that each of the columns goes to $0$. Then you want to know whether that guarantees that the row sums go to a finite limit. Am I correct?
    – saulspatz
    Aug 2 at 2:17











  • This cannot be answered in general without more specifics or, even better, some example(s). each element of the sequence is a function also of the length of the sequence This is always the case for Riemann sums for example, see e.g. How do you calculate this limit $lim_ntoinftysum_k=1^n frackn^2+k^2$?, but this is by no means a general answer.
    – dxiv
    Aug 2 at 2:42











  • @saulspatz Yes, good image.
    – Alecos Papadopoulos
    Aug 2 at 5:34












  • 1




    I just want to make sure I understand. I'm think of this as an infinite lower triangular matrix, where $S_n$ is the sum of the $n$th row. When you say that each element of the sequence goes to $0,$ it means that each of the columns goes to $0$. Then you want to know whether that guarantees that the row sums go to a finite limit. Am I correct?
    – saulspatz
    Aug 2 at 2:17











  • This cannot be answered in general without more specifics or, even better, some example(s). each element of the sequence is a function also of the length of the sequence This is always the case for Riemann sums for example, see e.g. How do you calculate this limit $lim_ntoinftysum_k=1^n frackn^2+k^2$?, but this is by no means a general answer.
    – dxiv
    Aug 2 at 2:42











  • @saulspatz Yes, good image.
    – Alecos Papadopoulos
    Aug 2 at 5:34







1




1




I just want to make sure I understand. I'm think of this as an infinite lower triangular matrix, where $S_n$ is the sum of the $n$th row. When you say that each element of the sequence goes to $0,$ it means that each of the columns goes to $0$. Then you want to know whether that guarantees that the row sums go to a finite limit. Am I correct?
– saulspatz
Aug 2 at 2:17





I just want to make sure I understand. I'm think of this as an infinite lower triangular matrix, where $S_n$ is the sum of the $n$th row. When you say that each element of the sequence goes to $0,$ it means that each of the columns goes to $0$. Then you want to know whether that guarantees that the row sums go to a finite limit. Am I correct?
– saulspatz
Aug 2 at 2:17













This cannot be answered in general without more specifics or, even better, some example(s). each element of the sequence is a function also of the length of the sequence This is always the case for Riemann sums for example, see e.g. How do you calculate this limit $lim_ntoinftysum_k=1^n frackn^2+k^2$?, but this is by no means a general answer.
– dxiv
Aug 2 at 2:42





This cannot be answered in general without more specifics or, even better, some example(s). each element of the sequence is a function also of the length of the sequence This is always the case for Riemann sums for example, see e.g. How do you calculate this limit $lim_ntoinftysum_k=1^n frackn^2+k^2$?, but this is by no means a general answer.
– dxiv
Aug 2 at 2:42













@saulspatz Yes, good image.
– Alecos Papadopoulos
Aug 2 at 5:34




@saulspatz Yes, good image.
– Alecos Papadopoulos
Aug 2 at 5:34










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If I've understood the question correctly, as indicated in my comment, then the answer is "no." For example set $$
b_i(n)=cases1over n,&$1leq ileq n, $if $n$ is odd\
2over n,&$1leq ileq n, $if $n$ is even
$$
so that $S_n=1$ if $n$ is odd, and $S_n=2$ is $n$ is even, and $S_n$ does not converge.



Or you could define $$b_i(n) =log nover n$$ to get $S_ntoinfty.$



EDIT



I just remembered that you want $b_i(n)$ to be rational. That makes no difference. Just take a good rational approximation to $log n.$






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    If I've understood the question correctly, as indicated in my comment, then the answer is "no." For example set $$
    b_i(n)=cases1over n,&$1leq ileq n, $if $n$ is odd\
    2over n,&$1leq ileq n, $if $n$ is even
    $$
    so that $S_n=1$ if $n$ is odd, and $S_n=2$ is $n$ is even, and $S_n$ does not converge.



    Or you could define $$b_i(n) =log nover n$$ to get $S_ntoinfty.$



    EDIT



    I just remembered that you want $b_i(n)$ to be rational. That makes no difference. Just take a good rational approximation to $log n.$






    share|cite|improve this answer

























      up vote
      1
      down vote













      If I've understood the question correctly, as indicated in my comment, then the answer is "no." For example set $$
      b_i(n)=cases1over n,&$1leq ileq n, $if $n$ is odd\
      2over n,&$1leq ileq n, $if $n$ is even
      $$
      so that $S_n=1$ if $n$ is odd, and $S_n=2$ is $n$ is even, and $S_n$ does not converge.



      Or you could define $$b_i(n) =log nover n$$ to get $S_ntoinfty.$



      EDIT



      I just remembered that you want $b_i(n)$ to be rational. That makes no difference. Just take a good rational approximation to $log n.$






      share|cite|improve this answer























        up vote
        1
        down vote










        up vote
        1
        down vote









        If I've understood the question correctly, as indicated in my comment, then the answer is "no." For example set $$
        b_i(n)=cases1over n,&$1leq ileq n, $if $n$ is odd\
        2over n,&$1leq ileq n, $if $n$ is even
        $$
        so that $S_n=1$ if $n$ is odd, and $S_n=2$ is $n$ is even, and $S_n$ does not converge.



        Or you could define $$b_i(n) =log nover n$$ to get $S_ntoinfty.$



        EDIT



        I just remembered that you want $b_i(n)$ to be rational. That makes no difference. Just take a good rational approximation to $log n.$






        share|cite|improve this answer













        If I've understood the question correctly, as indicated in my comment, then the answer is "no." For example set $$
        b_i(n)=cases1over n,&$1leq ileq n, $if $n$ is odd\
        2over n,&$1leq ileq n, $if $n$ is even
        $$
        so that $S_n=1$ if $n$ is odd, and $S_n=2$ is $n$ is even, and $S_n$ does not converge.



        Or you could define $$b_i(n) =log nover n$$ to get $S_ntoinfty.$



        EDIT



        I just remembered that you want $b_i(n)$ to be rational. That makes no difference. Just take a good rational approximation to $log n.$







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Aug 2 at 2:27









        saulspatz

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