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Series that evaluates to different values
What is the sum of the infinite telescopic series $sum_n=1^infty arctan (n+4)-arctan(n+2))$?Convergence of the series using power seriesSimplify this summation: $sumlimits_k=5^inftybinomk-1k-5frack^32^k$Convergent series with general term $a_nb_n$Decide whether the series $sum_n=1^infty frac1+5^n1+6^n$ converges or divergesCan different choices of regulator assign different values to the same divergent series?Riemann series theorem - when can I be sure that my series diverges?Find for which values of $x$ the series convergesWorking on series $A(x):=sum_n=0^infty2^-ncos(nx)$Expected value of an r.v. that counts the number of pickable values that never appear in a random sequence
$begingroup$
I am trying to find a sequence $x_pq$ such that $sum_p=1^inftysum_q=1^infty x_pq$ and $sum_q=1^inftysum_p=1^infty x_pq$ both exist and evaluate to different values. I am thinking of this as a matrix, but I can't seem to see how adding all the elements of it column wise vs row-wise would change the total sum? Any hints/examples would be helpful.
real-analysis sequences-and-series summation
$endgroup$
add a comment |
$begingroup$
I am trying to find a sequence $x_pq$ such that $sum_p=1^inftysum_q=1^infty x_pq$ and $sum_q=1^inftysum_p=1^infty x_pq$ both exist and evaluate to different values. I am thinking of this as a matrix, but I can't seem to see how adding all the elements of it column wise vs row-wise would change the total sum? Any hints/examples would be helpful.
real-analysis sequences-and-series summation
$endgroup$
1
$begingroup$
Is this the kind of thing you're looking for? The second paragraph gives an example and the convergent values for which a series can be rearranged and obtain different values.
$endgroup$
– Clayton
May 9 at 21:52
1
$begingroup$
This theorem is what motivates my question - but how would you write an explicit example with 2 indices?
$endgroup$
– user672611
May 9 at 22:00
$begingroup$
One of the classical example (which should be well-attested enough to find a detailed proof for) is the Eisenstein series $G_2(tau) = sum_m, n (n + m tau)^-2$. Unlike the higher $G_n$, swapping the order of the summation $n, m$ produces an extra $2pi i/tau$ term.
$endgroup$
– anomaly
May 10 at 2:41
add a comment |
$begingroup$
I am trying to find a sequence $x_pq$ such that $sum_p=1^inftysum_q=1^infty x_pq$ and $sum_q=1^inftysum_p=1^infty x_pq$ both exist and evaluate to different values. I am thinking of this as a matrix, but I can't seem to see how adding all the elements of it column wise vs row-wise would change the total sum? Any hints/examples would be helpful.
real-analysis sequences-and-series summation
$endgroup$
I am trying to find a sequence $x_pq$ such that $sum_p=1^inftysum_q=1^infty x_pq$ and $sum_q=1^inftysum_p=1^infty x_pq$ both exist and evaluate to different values. I am thinking of this as a matrix, but I can't seem to see how adding all the elements of it column wise vs row-wise would change the total sum? Any hints/examples would be helpful.
real-analysis sequences-and-series summation
real-analysis sequences-and-series summation
edited May 10 at 19:02
asked May 9 at 21:47
user672611
1
$begingroup$
Is this the kind of thing you're looking for? The second paragraph gives an example and the convergent values for which a series can be rearranged and obtain different values.
$endgroup$
– Clayton
May 9 at 21:52
1
$begingroup$
This theorem is what motivates my question - but how would you write an explicit example with 2 indices?
$endgroup$
– user672611
May 9 at 22:00
$begingroup$
One of the classical example (which should be well-attested enough to find a detailed proof for) is the Eisenstein series $G_2(tau) = sum_m, n (n + m tau)^-2$. Unlike the higher $G_n$, swapping the order of the summation $n, m$ produces an extra $2pi i/tau$ term.
$endgroup$
– anomaly
May 10 at 2:41
add a comment |
1
$begingroup$
Is this the kind of thing you're looking for? The second paragraph gives an example and the convergent values for which a series can be rearranged and obtain different values.
$endgroup$
– Clayton
May 9 at 21:52
1
$begingroup$
This theorem is what motivates my question - but how would you write an explicit example with 2 indices?
$endgroup$
– user672611
May 9 at 22:00
$begingroup$
One of the classical example (which should be well-attested enough to find a detailed proof for) is the Eisenstein series $G_2(tau) = sum_m, n (n + m tau)^-2$. Unlike the higher $G_n$, swapping the order of the summation $n, m$ produces an extra $2pi i/tau$ term.
$endgroup$
– anomaly
May 10 at 2:41
1
1
$begingroup$
Is this the kind of thing you're looking for? The second paragraph gives an example and the convergent values for which a series can be rearranged and obtain different values.
$endgroup$
– Clayton
May 9 at 21:52
$begingroup$
Is this the kind of thing you're looking for? The second paragraph gives an example and the convergent values for which a series can be rearranged and obtain different values.
$endgroup$
– Clayton
May 9 at 21:52
1
1
$begingroup$
This theorem is what motivates my question - but how would you write an explicit example with 2 indices?
$endgroup$
– user672611
May 9 at 22:00
$begingroup$
This theorem is what motivates my question - but how would you write an explicit example with 2 indices?
$endgroup$
– user672611
May 9 at 22:00
$begingroup$
One of the classical example (which should be well-attested enough to find a detailed proof for) is the Eisenstein series $G_2(tau) = sum_m, n (n + m tau)^-2$. Unlike the higher $G_n$, swapping the order of the summation $n, m$ produces an extra $2pi i/tau$ term.
$endgroup$
– anomaly
May 10 at 2:41
$begingroup$
One of the classical example (which should be well-attested enough to find a detailed proof for) is the Eisenstein series $G_2(tau) = sum_m, n (n + m tau)^-2$. Unlike the higher $G_n$, swapping the order of the summation $n, m$ produces an extra $2pi i/tau$ term.
$endgroup$
– anomaly
May 10 at 2:41
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Maybe you are looking for something like this: $$beginpmatrix1 &-1 & 0&0&0 & 0 &cdots \
0 & 1 & -1& 0 & 0 & 0&cdots\
0 & 0 & 1 & -1 & 0 & 0&cdots\
0 & 0 & 0 & 1 & -1 & 0&cdots \
0 & 0 & 0 & 0 & 1 & -1&cdots \
vdots &vdots &vdots & vdots & vdots & ddots&ddots
endpmatrix.$$
So that $x_p,p = 1$, $x_p,p+1=-1$ and $x_p,q = 0$ if $q neq p,p+1$.
Here we have $$sum_q=1^infty x_p,q = 0$$ for all $p in mathbb N$, and so $$sum^infty_p=1 sum^infty_q=1 x_p,q = 0.$$ However, we also have $$sum_p=1^infty x_p,q = 0 ,,,, text for ,,,, q ge 2, ,,,,, text but ,,,,, sum_p=1^infty x_p,1 = 1$$ and so $$sum^infty_q=1 sum^infty_p=1 x_p,q = 1.$$
As was said in the other answer, if $$sum_p,q lvert x_p,qrvert$$ converges (which isn't the case here), then Fubini's theorem tells us we can sum in either dimension first and get the same answer. Similarly, if all $x_p,q >0$, we can do the same, by Tonelli's theorem and this allows for the possibility that the sums evaluate to $+infty$.
edited May 10 at 0:50
Brian Borchers
6,55011320
answered May 9 at 22:51
User8128User8128
11.2k1622
$endgroup$
$begingroup$
Thats a good example where the sums are not just of different sign (+1).
$endgroup$
– RRL
May 9 at 22:53
add a comment |
$begingroup$
The iterated sums will be equal (if they converge) when all terms are nonnegative or, more generally, when we have absolute convergence.
Consider, on the other hand, $a_jk = 1/(k^2-j^2)$ if $jneq k$ and $a_jk = 0$ if $j=k$.
Here we have,
$$fracpi^28=tag*sum_j=1^infty sum_k=1\jneq k^infty frac1k^2 - j^2 neq sum_k=1^infty sum_j=1\j neq k^infty frac1k^2-j^2 = -fracpi^28 $$
To obtain the sum, note that
$$sum_k=1\jneq k^K frac1k^2-j^2 = frac12jsum_k=1\jneq k^K left(frac1k-j - frac1k+j right) \= frac12jleft(-sum_k=1^j-1frac1k + sum_k=1^K-jfrac1k - sum_k=j+1^K+jfrac1k + frac12j right)\= frac12jleft(frac1j +sum_k=j+1^K-jfrac1k - sum_k=j+1^K+jfrac1k + frac12j right)\ = frac12jleft(frac32j - sum_k=-j+1^j frac1K+kright) $$
and,
$$sum_k=1\jneq k^infty frac1k^2-j^2 = lim_K to inftyfrac12jleft(frac32j - sum_k=-n+1^n frac1K+kright) = frac34j^2$$
Thus,
$$sum_j=1^infty sum_k=1\jneq k^infty frac1k^2 - j^2 = sum_j=1^infty frac34j^2 = frac34 fracpi^26$$
answered May 9 at 22:47
RRLRRL
55k52776
$endgroup$
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
Maybe you are looking for something like this: $$beginpmatrix1 &-1 & 0&0&0 & 0 &cdots \
0 & 1 & -1& 0 & 0 & 0&cdots\
0 & 0 & 1 & -1 & 0 & 0&cdots\
0 & 0 & 0 & 1 & -1 & 0&cdots \
0 & 0 & 0 & 0 & 1 & -1&cdots \
vdots &vdots &vdots & vdots & vdots & ddots&ddots
endpmatrix.$$
So that $x_p,p = 1$, $x_p,p+1=-1$ and $x_p,q = 0$ if $q neq p,p+1$.
Here we have $$sum_q=1^infty x_p,q = 0$$ for all $p in mathbb N$, and so $$sum^infty_p=1 sum^infty_q=1 x_p,q = 0.$$ However, we also have $$sum_p=1^infty x_p,q = 0 ,,,, text for ,,,, q ge 2, ,,,,, text but ,,,,, sum_p=1^infty x_p,1 = 1$$ and so $$sum^infty_q=1 sum^infty_p=1 x_p,q = 1.$$
As was said in the other answer, if $$sum_p,q lvert x_p,qrvert$$ converges (which isn't the case here), then Fubini's theorem tells us we can sum in either dimension first and get the same answer. Similarly, if all $x_p,q >0$, we can do the same, by Tonelli's theorem and this allows for the possibility that the sums evaluate to $+infty$.
edited May 10 at 0:50
Brian Borchers
6,55011320
answered May 9 at 22:51
User8128User8128
11.2k1622
$endgroup$
$begingroup$
Thats a good example where the sums are not just of different sign (+1).
$endgroup$
– RRL
May 9 at 22:53
add a comment |
$begingroup$
Maybe you are looking for something like this: $$beginpmatrix1 &-1 & 0&0&0 & 0 &cdots \
0 & 1 & -1& 0 & 0 & 0&cdots\
0 & 0 & 1 & -1 & 0 & 0&cdots\
0 & 0 & 0 & 1 & -1 & 0&cdots \
0 & 0 & 0 & 0 & 1 & -1&cdots \
vdots &vdots &vdots & vdots & vdots & ddots&ddots
endpmatrix.$$
So that $x_p,p = 1$, $x_p,p+1=-1$ and $x_p,q = 0$ if $q neq p,p+1$.
Here we have $$sum_q=1^infty x_p,q = 0$$ for all $p in mathbb N$, and so $$sum^infty_p=1 sum^infty_q=1 x_p,q = 0.$$ However, we also have $$sum_p=1^infty x_p,q = 0 ,,,, text for ,,,, q ge 2, ,,,,, text but ,,,,, sum_p=1^infty x_p,1 = 1$$ and so $$sum^infty_q=1 sum^infty_p=1 x_p,q = 1.$$
As was said in the other answer, if $$sum_p,q lvert x_p,qrvert$$ converges (which isn't the case here), then Fubini's theorem tells us we can sum in either dimension first and get the same answer. Similarly, if all $x_p,q >0$, we can do the same, by Tonelli's theorem and this allows for the possibility that the sums evaluate to $+infty$.
edited May 10 at 0:50
Brian Borchers
6,55011320
answered May 9 at 22:51
User8128User8128
11.2k1622
$endgroup$
$begingroup$
Thats a good example where the sums are not just of different sign (+1).
$endgroup$
– RRL
May 9 at 22:53
add a comment |
$begingroup$
Maybe you are looking for something like this: $$beginpmatrix1 &-1 & 0&0&0 & 0 &cdots \
0 & 1 & -1& 0 & 0 & 0&cdots\
0 & 0 & 1 & -1 & 0 & 0&cdots\
0 & 0 & 0 & 1 & -1 & 0&cdots \
0 & 0 & 0 & 0 & 1 & -1&cdots \
vdots &vdots &vdots & vdots & vdots & ddots&ddots
endpmatrix.$$
So that $x_p,p = 1$, $x_p,p+1=-1$ and $x_p,q = 0$ if $q neq p,p+1$.
Here we have $$sum_q=1^infty x_p,q = 0$$ for all $p in mathbb N$, and so $$sum^infty_p=1 sum^infty_q=1 x_p,q = 0.$$ However, we also have $$sum_p=1^infty x_p,q = 0 ,,,, text for ,,,, q ge 2, ,,,,, text but ,,,,, sum_p=1^infty x_p,1 = 1$$ and so $$sum^infty_q=1 sum^infty_p=1 x_p,q = 1.$$
As was said in the other answer, if $$sum_p,q lvert x_p,qrvert$$ converges (which isn't the case here), then Fubini's theorem tells us we can sum in either dimension first and get the same answer. Similarly, if all $x_p,q >0$, we can do the same, by Tonelli's theorem and this allows for the possibility that the sums evaluate to $+infty$.
edited May 10 at 0:50
Brian Borchers
6,55011320
answered May 9 at 22:51
User8128User8128
11.2k1622
$endgroup$
Maybe you are looking for something like this: $$beginpmatrix1 &-1 & 0&0&0 & 0 &cdots \
0 & 1 & -1& 0 & 0 & 0&cdots\
0 & 0 & 1 & -1 & 0 & 0&cdots\
0 & 0 & 0 & 1 & -1 & 0&cdots \
0 & 0 & 0 & 0 & 1 & -1&cdots \
vdots &vdots &vdots & vdots & vdots & ddots&ddots
endpmatrix.$$
So that $x_p,p = 1$, $x_p,p+1=-1$ and $x_p,q = 0$ if $q neq p,p+1$.
Here we have $$sum_q=1^infty x_p,q = 0$$ for all $p in mathbb N$, and so $$sum^infty_p=1 sum^infty_q=1 x_p,q = 0.$$ However, we also have $$sum_p=1^infty x_p,q = 0 ,,,, text for ,,,, q ge 2, ,,,,, text but ,,,,, sum_p=1^infty x_p,1 = 1$$ and so $$sum^infty_q=1 sum^infty_p=1 x_p,q = 1.$$
As was said in the other answer, if $$sum_p,q lvert x_p,qrvert$$ converges (which isn't the case here), then Fubini's theorem tells us we can sum in either dimension first and get the same answer. Similarly, if all $x_p,q >0$, we can do the same, by Tonelli's theorem and this allows for the possibility that the sums evaluate to $+infty$.
edited May 10 at 0:50
Brian Borchers
6,55011320
answered May 9 at 22:51
User8128User8128
11.2k1622
edited May 10 at 0:50
Brian Borchers
6,55011320
edited May 10 at 0:50
Brian Borchers
6,55011320
edited May 10 at 0:50
Brian Borchers
6,55011320
6,55011320
answered May 9 at 22:51
User8128User8128
11.2k1622
answered May 9 at 22:51
User8128User8128
11.2k1622
answered May 9 at 22:51
User8128User8128
11.2k1622
11.2k1622
$begingroup$
Thats a good example where the sums are not just of different sign (+1).
$endgroup$
– RRL
May 9 at 22:53
add a comment |
$begingroup$
Thats a good example where the sums are not just of different sign (+1).
$endgroup$
– RRL
May 9 at 22:53
$begingroup$
Thats a good example where the sums are not just of different sign (+1).
$endgroup$
– RRL
May 9 at 22:53
$begingroup$
Thats a good example where the sums are not just of different sign (+1).
$endgroup$
– RRL
May 9 at 22:53
add a comment |
$begingroup$
The iterated sums will be equal (if they converge) when all terms are nonnegative or, more generally, when we have absolute convergence.
Consider, on the other hand, $a_jk = 1/(k^2-j^2)$ if $jneq k$ and $a_jk = 0$ if $j=k$.
Here we have,
$$fracpi^28=tag*sum_j=1^infty sum_k=1\jneq k^infty frac1k^2 - j^2 neq sum_k=1^infty sum_j=1\j neq k^infty frac1k^2-j^2 = -fracpi^28 $$
To obtain the sum, note that
$$sum_k=1\jneq k^K frac1k^2-j^2 = frac12jsum_k=1\jneq k^K left(frac1k-j - frac1k+j right) \= frac12jleft(-sum_k=1^j-1frac1k + sum_k=1^K-jfrac1k - sum_k=j+1^K+jfrac1k + frac12j right)\= frac12jleft(frac1j +sum_k=j+1^K-jfrac1k - sum_k=j+1^K+jfrac1k + frac12j right)\ = frac12jleft(frac32j - sum_k=-j+1^j frac1K+kright) $$
and,
$$sum_k=1\jneq k^infty frac1k^2-j^2 = lim_K to inftyfrac12jleft(frac32j - sum_k=-n+1^n frac1K+kright) = frac34j^2$$
Thus,
$$sum_j=1^infty sum_k=1\jneq k^infty frac1k^2 - j^2 = sum_j=1^infty frac34j^2 = frac34 fracpi^26$$
answered May 9 at 22:47
RRLRRL
55k52776
$endgroup$
add a comment |
$begingroup$
The iterated sums will be equal (if they converge) when all terms are nonnegative or, more generally, when we have absolute convergence.
Consider, on the other hand, $a_jk = 1/(k^2-j^2)$ if $jneq k$ and $a_jk = 0$ if $j=k$.
Here we have,
$$fracpi^28=tag*sum_j=1^infty sum_k=1\jneq k^infty frac1k^2 - j^2 neq sum_k=1^infty sum_j=1\j neq k^infty frac1k^2-j^2 = -fracpi^28 $$
To obtain the sum, note that
$$sum_k=1\jneq k^K frac1k^2-j^2 = frac12jsum_k=1\jneq k^K left(frac1k-j - frac1k+j right) \= frac12jleft(-sum_k=1^j-1frac1k + sum_k=1^K-jfrac1k - sum_k=j+1^K+jfrac1k + frac12j right)\= frac12jleft(frac1j +sum_k=j+1^K-jfrac1k - sum_k=j+1^K+jfrac1k + frac12j right)\ = frac12jleft(frac32j - sum_k=-j+1^j frac1K+kright) $$
and,
$$sum_k=1\jneq k^infty frac1k^2-j^2 = lim_K to inftyfrac12jleft(frac32j - sum_k=-n+1^n frac1K+kright) = frac34j^2$$
Thus,
$$sum_j=1^infty sum_k=1\jneq k^infty frac1k^2 - j^2 = sum_j=1^infty frac34j^2 = frac34 fracpi^26$$
answered May 9 at 22:47
RRLRRL
55k52776
$endgroup$
add a comment |
$begingroup$
The iterated sums will be equal (if they converge) when all terms are nonnegative or, more generally, when we have absolute convergence.
Consider, on the other hand, $a_jk = 1/(k^2-j^2)$ if $jneq k$ and $a_jk = 0$ if $j=k$.
Here we have,
$$fracpi^28=tag*sum_j=1^infty sum_k=1\jneq k^infty frac1k^2 - j^2 neq sum_k=1^infty sum_j=1\j neq k^infty frac1k^2-j^2 = -fracpi^28 $$
To obtain the sum, note that
$$sum_k=1\jneq k^K frac1k^2-j^2 = frac12jsum_k=1\jneq k^K left(frac1k-j - frac1k+j right) \= frac12jleft(-sum_k=1^j-1frac1k + sum_k=1^K-jfrac1k - sum_k=j+1^K+jfrac1k + frac12j right)\= frac12jleft(frac1j +sum_k=j+1^K-jfrac1k - sum_k=j+1^K+jfrac1k + frac12j right)\ = frac12jleft(frac32j - sum_k=-j+1^j frac1K+kright) $$
and,
$$sum_k=1\jneq k^infty frac1k^2-j^2 = lim_K to inftyfrac12jleft(frac32j - sum_k=-n+1^n frac1K+kright) = frac34j^2$$
Thus,
$$sum_j=1^infty sum_k=1\jneq k^infty frac1k^2 - j^2 = sum_j=1^infty frac34j^2 = frac34 fracpi^26$$
answered May 9 at 22:47
RRLRRL
55k52776
$endgroup$
The iterated sums will be equal (if they converge) when all terms are nonnegative or, more generally, when we have absolute convergence.
Consider, on the other hand, $a_jk = 1/(k^2-j^2)$ if $jneq k$ and $a_jk = 0$ if $j=k$.
Here we have,
$$fracpi^28=tag*sum_j=1^infty sum_k=1\jneq k^infty frac1k^2 - j^2 neq sum_k=1^infty sum_j=1\j neq k^infty frac1k^2-j^2 = -fracpi^28 $$
To obtain the sum, note that
$$sum_k=1\jneq k^K frac1k^2-j^2 = frac12jsum_k=1\jneq k^K left(frac1k-j - frac1k+j right) \= frac12jleft(-sum_k=1^j-1frac1k + sum_k=1^K-jfrac1k - sum_k=j+1^K+jfrac1k + frac12j right)\= frac12jleft(frac1j +sum_k=j+1^K-jfrac1k - sum_k=j+1^K+jfrac1k + frac12j right)\ = frac12jleft(frac32j - sum_k=-j+1^j frac1K+kright) $$
and,
$$sum_k=1\jneq k^infty frac1k^2-j^2 = lim_K to inftyfrac12jleft(frac32j - sum_k=-n+1^n frac1K+kright) = frac34j^2$$
Thus,
$$sum_j=1^infty sum_k=1\jneq k^infty frac1k^2 - j^2 = sum_j=1^infty frac34j^2 = frac34 fracpi^26$$
answered May 9 at 22:47
RRLRRL
55k52776
answered May 9 at 22:47
RRLRRL
55k52776
answered May 9 at 22:47
RRLRRL
55k52776
answered May 9 at 22:47
RRLRRL
55k52776
55k52776
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1
$begingroup$
Is this the kind of thing you're looking for? The second paragraph gives an example and the convergent values for which a series can be rearranged and obtain different values.
$endgroup$
– Clayton
May 9 at 21:52
1
$begingroup$
This theorem is what motivates my question - but how would you write an explicit example with 2 indices?
$endgroup$
– user672611
May 9 at 22:00
$begingroup$
One of the classical example (which should be well-attested enough to find a detailed proof for) is the Eisenstein series $G_2(tau) = sum_m, n (n + m tau)^-2$. Unlike the higher $G_n$, swapping the order of the summation $n, m$ produces an extra $2pi i/tau$ term.
$endgroup$
– anomaly
May 10 at 2:41