Asymptotic behavior of series tail












1












$begingroup$


Suppose the series $sum_{n=1}^infty n a_n$ converges. Then I would like show (if it is always true):



$$lim_{N to infty} N sum_{n= N}^infty a_n = 0.$$



My work:



I started with the condition $a_n > 0$ for all $n$. I know that $sum_{n=1}^infty na_n$ converges and therefore $lim_{N to infty}sum_{n=N}^infty na_n = 0$.



Since $Na_n leq na_n$ for $n geq N$ it holds that $Nsum_{n=N}^infty a_n leq sum_{n=N}^infty n a_n $ and therefore



$$0 leq lim_{N to infty}Nsum_{n=N}^infty a_n leq lim_{N to infty}sum_{n=N}^infty na_n = 0 $$



But for a general sequence${a_n}$ that is not always or eventually nonnegative or nonpositive is this still true?



I suspect it is not but could not find a counterexample.






sequences-and-series







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Summation by parts.
    $endgroup$
    – Jack D'Aurizio
    Jan 15 at 22:00
















1












$begingroup$


Suppose the series $sum_{n=1}^infty n a_n$ converges. Then I would like show (if it is always true):



$$lim_{N to infty} N sum_{n= N}^infty a_n = 0.$$



My work:



I started with the condition $a_n > 0$ for all $n$. I know that $sum_{n=1}^infty na_n$ converges and therefore $lim_{N to infty}sum_{n=N}^infty na_n = 0$.



Since $Na_n leq na_n$ for $n geq N$ it holds that $Nsum_{n=N}^infty a_n leq sum_{n=N}^infty n a_n $ and therefore



$$0 leq lim_{N to infty}Nsum_{n=N}^infty a_n leq lim_{N to infty}sum_{n=N}^infty na_n = 0 $$



But for a general sequence${a_n}$ that is not always or eventually nonnegative or nonpositive is this still true?



I suspect it is not but could not find a counterexample.






sequences-and-series







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Summation by parts.
    $endgroup$
    – Jack D'Aurizio
    Jan 15 at 22:00














1












1








1





$begingroup$


Suppose the series $sum_{n=1}^infty n a_n$ converges. Then I would like show (if it is always true):



$$lim_{N to infty} N sum_{n= N}^infty a_n = 0.$$



My work:



I started with the condition $a_n > 0$ for all $n$. I know that $sum_{n=1}^infty na_n$ converges and therefore $lim_{N to infty}sum_{n=N}^infty na_n = 0$.



Since $Na_n leq na_n$ for $n geq N$ it holds that $Nsum_{n=N}^infty a_n leq sum_{n=N}^infty n a_n $ and therefore



$$0 leq lim_{N to infty}Nsum_{n=N}^infty a_n leq lim_{N to infty}sum_{n=N}^infty na_n = 0 $$



But for a general sequence${a_n}$ that is not always or eventually nonnegative or nonpositive is this still true?



I suspect it is not but could not find a counterexample.






sequences-and-series







share|cite|improve this question









$endgroup$




Suppose the series $sum_{n=1}^infty n a_n$ converges. Then I would like show (if it is always true):



$$lim_{N to infty} N sum_{n= N}^infty a_n = 0.$$



My work:



I started with the condition $a_n > 0$ for all $n$. I know that $sum_{n=1}^infty na_n$ converges and therefore $lim_{N to infty}sum_{n=N}^infty na_n = 0$.



Since $Na_n leq na_n$ for $n geq N$ it holds that $Nsum_{n=N}^infty a_n leq sum_{n=N}^infty n a_n $ and therefore



$$0 leq lim_{N to infty}Nsum_{n=N}^infty a_n leq lim_{N to infty}sum_{n=N}^infty na_n = 0 $$



But for a general sequence${a_n}$ that is not always or eventually nonnegative or nonpositive is this still true?



I suspect it is not but could not find a counterexample.






sequences-and-series




sequences-and-series






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 15 at 21:56









SASSAS

406310




406310








  • 1




    $begingroup$
    Summation by parts.
    $endgroup$
    – Jack D'Aurizio
    Jan 15 at 22:00














  • 1




    $begingroup$
    Summation by parts.
    $endgroup$
    – Jack D'Aurizio
    Jan 15 at 22:00








1




1




$begingroup$
Summation by parts.
$endgroup$
– Jack D'Aurizio
Jan 15 at 22:00




$begingroup$
Summation by parts.
$endgroup$
– Jack D'Aurizio
Jan 15 at 22:00










1 Answer
1






active

oldest

votes


















1












$begingroup$

The sign of $a_n$ is not relevant, although it is somewhat harder to prove the result.



Assume there exists a finite number $S$ such that $S_N = sum_{n=1}^N n a_n to S$ as $N to infty$. Using summation by parts we have



$$ Nsum_{n=N}^M a_n = Nsum_{n=N}^M n a_n frac{1}{n} = Nleft[frac{S_M}{M} - frac{S_{N-1}}{N} + sum_{n=N}^{M-1} S_nleft(frac{1}{n} - frac{1}{n+1} right)right]$$



Taking the limit as $M to infty$ we get



$$Nsum_{n=N}^infty a_n = -S_{N-1} + Nsum_{n=N}^{infty} S_nleft(frac{1}{n} - frac{1}{n+1} right)$$



Since $S- epsilon < S_n < S+ epsilon$ for sufficiently large $n$ , it can be shown that the limit of the sum on the RHS is $S$ and, thus,



$$lim_{Nto infty}Nsum_{n=N}^infty a_n = -S + S = 0$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you. So that I understand fully you are saying $ S- epsilon < N sum_{n=N}^infty S_n(1/n - 1/(n+1)) <S+ epsilon$ because $N sum_{n=N}^infty(1/n - 1/(n+1)) = N(1/N) = 1$ and $S- epsilon < S_n < S + epsilon$ for $n geq N$?
    $endgroup$
    – SAS
    Jan 16 at 0:11










  • $begingroup$
    @SAS: Yes -- that is correct.
    $endgroup$
    – RRL
    Jan 16 at 4:35












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1 Answer
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1












$begingroup$

The sign of $a_n$ is not relevant, although it is somewhat harder to prove the result.



Assume there exists a finite number $S$ such that $S_N = sum_{n=1}^N n a_n to S$ as $N to infty$. Using summation by parts we have



$$ Nsum_{n=N}^M a_n = Nsum_{n=N}^M n a_n frac{1}{n} = Nleft[frac{S_M}{M} - frac{S_{N-1}}{N} + sum_{n=N}^{M-1} S_nleft(frac{1}{n} - frac{1}{n+1} right)right]$$



Taking the limit as $M to infty$ we get



$$Nsum_{n=N}^infty a_n = -S_{N-1} + Nsum_{n=N}^{infty} S_nleft(frac{1}{n} - frac{1}{n+1} right)$$



Since $S- epsilon < S_n < S+ epsilon$ for sufficiently large $n$ , it can be shown that the limit of the sum on the RHS is $S$ and, thus,



$$lim_{Nto infty}Nsum_{n=N}^infty a_n = -S + S = 0$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you. So that I understand fully you are saying $ S- epsilon < N sum_{n=N}^infty S_n(1/n - 1/(n+1)) <S+ epsilon$ because $N sum_{n=N}^infty(1/n - 1/(n+1)) = N(1/N) = 1$ and $S- epsilon < S_n < S + epsilon$ for $n geq N$?
    $endgroup$
    – SAS
    Jan 16 at 0:11










  • $begingroup$
    @SAS: Yes -- that is correct.
    $endgroup$
    – RRL
    Jan 16 at 4:35
















1












$begingroup$

The sign of $a_n$ is not relevant, although it is somewhat harder to prove the result.



Assume there exists a finite number $S$ such that $S_N = sum_{n=1}^N n a_n to S$ as $N to infty$. Using summation by parts we have



$$ Nsum_{n=N}^M a_n = Nsum_{n=N}^M n a_n frac{1}{n} = Nleft[frac{S_M}{M} - frac{S_{N-1}}{N} + sum_{n=N}^{M-1} S_nleft(frac{1}{n} - frac{1}{n+1} right)right]$$



Taking the limit as $M to infty$ we get



$$Nsum_{n=N}^infty a_n = -S_{N-1} + Nsum_{n=N}^{infty} S_nleft(frac{1}{n} - frac{1}{n+1} right)$$



Since $S- epsilon < S_n < S+ epsilon$ for sufficiently large $n$ , it can be shown that the limit of the sum on the RHS is $S$ and, thus,



$$lim_{Nto infty}Nsum_{n=N}^infty a_n = -S + S = 0$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you. So that I understand fully you are saying $ S- epsilon < N sum_{n=N}^infty S_n(1/n - 1/(n+1)) <S+ epsilon$ because $N sum_{n=N}^infty(1/n - 1/(n+1)) = N(1/N) = 1$ and $S- epsilon < S_n < S + epsilon$ for $n geq N$?
    $endgroup$
    – SAS
    Jan 16 at 0:11










  • $begingroup$
    @SAS: Yes -- that is correct.
    $endgroup$
    – RRL
    Jan 16 at 4:35














1












1








1





$begingroup$

The sign of $a_n$ is not relevant, although it is somewhat harder to prove the result.



Assume there exists a finite number $S$ such that $S_N = sum_{n=1}^N n a_n to S$ as $N to infty$. Using summation by parts we have



$$ Nsum_{n=N}^M a_n = Nsum_{n=N}^M n a_n frac{1}{n} = Nleft[frac{S_M}{M} - frac{S_{N-1}}{N} + sum_{n=N}^{M-1} S_nleft(frac{1}{n} - frac{1}{n+1} right)right]$$



Taking the limit as $M to infty$ we get



$$Nsum_{n=N}^infty a_n = -S_{N-1} + Nsum_{n=N}^{infty} S_nleft(frac{1}{n} - frac{1}{n+1} right)$$



Since $S- epsilon < S_n < S+ epsilon$ for sufficiently large $n$ , it can be shown that the limit of the sum on the RHS is $S$ and, thus,



$$lim_{Nto infty}Nsum_{n=N}^infty a_n = -S + S = 0$$






share|cite|improve this answer









$endgroup$



The sign of $a_n$ is not relevant, although it is somewhat harder to prove the result.



Assume there exists a finite number $S$ such that $S_N = sum_{n=1}^N n a_n to S$ as $N to infty$. Using summation by parts we have



$$ Nsum_{n=N}^M a_n = Nsum_{n=N}^M n a_n frac{1}{n} = Nleft[frac{S_M}{M} - frac{S_{N-1}}{N} + sum_{n=N}^{M-1} S_nleft(frac{1}{n} - frac{1}{n+1} right)right]$$



Taking the limit as $M to infty$ we get



$$Nsum_{n=N}^infty a_n = -S_{N-1} + Nsum_{n=N}^{infty} S_nleft(frac{1}{n} - frac{1}{n+1} right)$$



Since $S- epsilon < S_n < S+ epsilon$ for sufficiently large $n$ , it can be shown that the limit of the sum on the RHS is $S$ and, thus,



$$lim_{Nto infty}Nsum_{n=N}^infty a_n = -S + S = 0$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 15 at 22:08









RRLRRL

54k52675




54k52675












  • $begingroup$
    Thank you. So that I understand fully you are saying $ S- epsilon < N sum_{n=N}^infty S_n(1/n - 1/(n+1)) <S+ epsilon$ because $N sum_{n=N}^infty(1/n - 1/(n+1)) = N(1/N) = 1$ and $S- epsilon < S_n < S + epsilon$ for $n geq N$?
    $endgroup$
    – SAS
    Jan 16 at 0:11










  • $begingroup$
    @SAS: Yes -- that is correct.
    $endgroup$
    – RRL
    Jan 16 at 4:35


















  • $begingroup$
    Thank you. So that I understand fully you are saying $ S- epsilon < N sum_{n=N}^infty S_n(1/n - 1/(n+1)) <S+ epsilon$ because $N sum_{n=N}^infty(1/n - 1/(n+1)) = N(1/N) = 1$ and $S- epsilon < S_n < S + epsilon$ for $n geq N$?
    $endgroup$
    – SAS
    Jan 16 at 0:11










  • $begingroup$
    @SAS: Yes -- that is correct.
    $endgroup$
    – RRL
    Jan 16 at 4:35
















$begingroup$
Thank you. So that I understand fully you are saying $ S- epsilon < N sum_{n=N}^infty S_n(1/n - 1/(n+1)) <S+ epsilon$ because $N sum_{n=N}^infty(1/n - 1/(n+1)) = N(1/N) = 1$ and $S- epsilon < S_n < S + epsilon$ for $n geq N$?
$endgroup$
– SAS
Jan 16 at 0:11




$begingroup$
Thank you. So that I understand fully you are saying $ S- epsilon < N sum_{n=N}^infty S_n(1/n - 1/(n+1)) <S+ epsilon$ because $N sum_{n=N}^infty(1/n - 1/(n+1)) = N(1/N) = 1$ and $S- epsilon < S_n < S + epsilon$ for $n geq N$?
$endgroup$
– SAS
Jan 16 at 0:11












$begingroup$
@SAS: Yes -- that is correct.
$endgroup$
– RRL
Jan 16 at 4:35




$begingroup$
@SAS: Yes -- that is correct.
$endgroup$
– RRL
Jan 16 at 4:35


















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