Asymptotic behavior of series tail
$begingroup$
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
asked Jan 15 at 21:56
SASSAS
406310
$endgroup$
add a comment |
$begingroup$
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
asked Jan 15 at 21:56
SASSAS
406310
$endgroup$
1
$begingroup$
Summation by parts.
$endgroup$
– Jack D'Aurizio
Jan 15 at 22:00
add a comment |
$begingroup$
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
asked Jan 15 at 21:56
SASSAS
406310
$endgroup$
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
asked Jan 15 at 21:56
SASSAS
406310
asked Jan 15 at 21:56
SASSAS
406310
asked Jan 15 at 21:56
SASSAS
406310
asked Jan 15 at 21:56
SASSAS
406310
asked Jan 15 at 21:56
SASSAS
406310
406310
1
$begingroup$
Summation by parts.
$endgroup$
– Jack D'Aurizio
Jan 15 at 22:00
add a comment |
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
add a comment |
1 Answer
1
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oldest
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$begingroup$
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,
$$lim_{Nto infty}Nsum_{n=N}^infty a_n = -S + S = 0$$
answered Jan 15 at 22:08
RRLRRL
54k52675
$endgroup$
$begingroup$
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
add a comment |
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$begingroup$
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,
$$lim_{Nto infty}Nsum_{n=N}^infty a_n = -S + S = 0$$
answered Jan 15 at 22:08
RRLRRL
54k52675
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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,
$$lim_{Nto infty}Nsum_{n=N}^infty a_n = -S + S = 0$$
answered Jan 15 at 22:08
RRLRRL
54k52675
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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,
$$lim_{Nto infty}Nsum_{n=N}^infty a_n = -S + S = 0$$
answered Jan 15 at 22:08
RRLRRL
54k52675
$endgroup$
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,
$$lim_{Nto infty}Nsum_{n=N}^infty a_n = -S + S = 0$$
answered Jan 15 at 22:08
RRLRRL
54k52675
answered Jan 15 at 22:08
RRLRRL
54k52675
answered Jan 15 at 22:08
RRLRRL
54k52675
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
Summation by parts.
$endgroup$
– Jack D'Aurizio
Jan 15 at 22:00