Mixing
Divergent Double Series $sum_i=1^Isum_j=1^Jfrac1i^2 + j^2 > int_1^I+1int_1^J+1 fracdxdyx^2 + y^2$.
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
For showing the double series $sum_i,j in mathbbZ^+frac1i^2 + j^2$ diverges, I can compare the partial sums with the double integral:
$$sum_i=1^Isum_j=1^Jfrac1i^2 + j^2 > int_1^I+1int_1^J+1 fracdxdyx^2 + y^2$$
and take limit of the integral as $I,J to infty$. I know that $dxdy/(x^2+y^2)$ behaves like $rdrdtheta/r^2$ in polar coordinates so the integral should diverge like $log(r)$ as $r to infty$.
I'm having trouble seeing how the region $[1,I+1]times[1,J+1]$ is transformed in polar coordinates and I'd like to see a rigorous proof that the integral diverges.
Thank you.
integration sequences-and-series
edited Aug 2 at 2:47
user 108128
18.8k41544
asked Aug 1 at 23:07
user28763
523
add a comment |Â
up vote
2
down vote
favorite
For showing the double series $sum_i,j in mathbbZ^+frac1i^2 + j^2$ diverges, I can compare the partial sums with the double integral:
$$sum_i=1^Isum_j=1^Jfrac1i^2 + j^2 > int_1^I+1int_1^J+1 fracdxdyx^2 + y^2$$
and take limit of the integral as $I,J to infty$. I know that $dxdy/(x^2+y^2)$ behaves like $rdrdtheta/r^2$ in polar coordinates so the integral should diverge like $log(r)$ as $r to infty$.
I'm having trouble seeing how the region $[1,I+1]times[1,J+1]$ is transformed in polar coordinates and I'd like to see a rigorous proof that the integral diverges.
Thank you.
integration sequences-and-series
edited Aug 2 at 2:47
user 108128
18.8k41544
asked Aug 1 at 23:07
user28763
523
2
Out of curiosity, why not use the fact that $arctan'(x) = frac11+x^2$ instead of switching to polar coordinates?
– Clement C.
Aug 1 at 23:12
1
Hint: the limit as $I,Jto infty$ is the integral over $[1, infty) times [1, infty)$ - and that contains, for example, the region $fracpi6 le theta le fracpi3, r ge 2$.
– Daniel Schepler
Aug 2 at 0:47
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
For showing the double series $sum_i,j in mathbbZ^+frac1i^2 + j^2$ diverges, I can compare the partial sums with the double integral:
$$sum_i=1^Isum_j=1^Jfrac1i^2 + j^2 > int_1^I+1int_1^J+1 fracdxdyx^2 + y^2$$
and take limit of the integral as $I,J to infty$. I know that $dxdy/(x^2+y^2)$ behaves like $rdrdtheta/r^2$ in polar coordinates so the integral should diverge like $log(r)$ as $r to infty$.
I'm having trouble seeing how the region $[1,I+1]times[1,J+1]$ is transformed in polar coordinates and I'd like to see a rigorous proof that the integral diverges.
Thank you.
integration sequences-and-series
edited Aug 2 at 2:47
user 108128
18.8k41544
asked Aug 1 at 23:07
user28763
523
For showing the double series $sum_i,j in mathbbZ^+frac1i^2 + j^2$ diverges, I can compare the partial sums with the double integral:
$$sum_i=1^Isum_j=1^Jfrac1i^2 + j^2 > int_1^I+1int_1^J+1 fracdxdyx^2 + y^2$$
and take limit of the integral as $I,J to infty$. I know that $dxdy/(x^2+y^2)$ behaves like $rdrdtheta/r^2$ in polar coordinates so the integral should diverge like $log(r)$ as $r to infty$.
I'm having trouble seeing how the region $[1,I+1]times[1,J+1]$ is transformed in polar coordinates and I'd like to see a rigorous proof that the integral diverges.
Thank you.
integration sequences-and-series
edited Aug 2 at 2:47
user 108128
18.8k41544
asked Aug 1 at 23:07
user28763
523
edited Aug 2 at 2:47
user 108128
18.8k41544
edited Aug 2 at 2:47
user 108128
18.8k41544
edited Aug 2 at 2:47
user 108128
18.8k41544
18.8k41544
asked Aug 1 at 23:07
user28763
523
asked Aug 1 at 23:07
user28763
523
asked Aug 1 at 23:07
user28763
523
523
2
Out of curiosity, why not use the fact that $arctan'(x) = frac11+x^2$ instead of switching to polar coordinates?
– Clement C.
Aug 1 at 23:12
1
Hint: the limit as $I,Jto infty$ is the integral over $[1, infty) times [1, infty)$ - and that contains, for example, the region $fracpi6 le theta le fracpi3, r ge 2$.
– Daniel Schepler
Aug 2 at 0:47
add a comment |Â
2
Out of curiosity, why not use the fact that $arctan'(x) = frac11+x^2$ instead of switching to polar coordinates?
– Clement C.
Aug 1 at 23:12
1
Hint: the limit as $I,Jto infty$ is the integral over $[1, infty) times [1, infty)$ - and that contains, for example, the region $fracpi6 le theta le fracpi3, r ge 2$.
– Daniel Schepler
Aug 2 at 0:47
2
2
Out of curiosity, why not use the fact that $arctan'(x) = frac11+x^2$ instead of switching to polar coordinates?
– Clement C.
Aug 1 at 23:12
Out of curiosity, why not use the fact that $arctan'(x) = frac11+x^2$ instead of switching to polar coordinates?
– Clement C.
Aug 1 at 23:12
1
1
Hint: the limit as $I,Jto infty$ is the integral over $[1, infty) times [1, infty)$ - and that contains, for example, the region $fracpi6 le theta le fracpi3, r ge 2$.
– Daniel Schepler
Aug 2 at 0:47
Hint: the limit as $I,Jto infty$ is the integral over $[1, infty) times [1, infty)$ - and that contains, for example, the region $fracpi6 le theta le fracpi3, r ge 2$.
– Daniel Schepler
Aug 2 at 0:47
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Assuming WLOG that $J leqslant I$ and changing variables as $x = 1+s, , y = 1+t$, we have
$$beginalign A_I,J=int_1^I+1int_1^J+1 fracdx , dyx^2 + y^2endalign geqslant int_0^Jint_0^J fracds , dts^2 + t^2 + 2(s+t) + 2$$
Now switch to polar coordinates $s = r cos theta, , t = r sin theta $.
The sector $S = (r,theta): 0 leqslant r leqslant J, ,0 leqslant theta leqslant pi/2$ is a subset of the square $[0,J]^2$. Since the integrand is nonnegative and $1 leqslant cos theta + sin theta leqslant sqrt2$ for $0 leqslant theta leqslant pi/2$ we have
$$beginalignA_I,J &geqslant int_0^pi/2int_0^J fracr, dr, dthetar^2 + 2r(cos theta + sin theta)+2\ &geqslant int_0^pi/2int_0^J fracr, dr, dthetar^2 + 2sqrt2r+2 \ &= fracpi2int_0^Jfracr(r+sqrt2)^2 , dr \ &=fracpi2int_0^Jfrac1r+sqrt2 , dr - fracpi2int_0^Jfracsqrt2(r+sqrt2)^2 , dr \ &= fracpi2left(log(J + sqrt2) + fracsqrt2J + sqrt2 - (1 + log sqrt2) right)endalign$$
The RHS tends to $+infty$ as $I,J to infty$.
answered Aug 3 at 17:09
RRL
43.4k42260
Thank you. These are the details I was interested in seeing.
– user28763
Aug 3 at 17:22
add a comment |Â
up vote
0
down vote
$int_1^I+1int_1^J+1fracdxdyx^2+y^2gt fracpi2int_sqrt2^min(I,J)+1fracdrr$ which becomes infinite like $log(min(I,J)+1)$ as both indicies become infinite.
answered Aug 2 at 0:31
herb steinberg
93529
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Assuming WLOG that $J leqslant I$ and changing variables as $x = 1+s, , y = 1+t$, we have
$$beginalign A_I,J=int_1^I+1int_1^J+1 fracdx , dyx^2 + y^2endalign geqslant int_0^Jint_0^J fracds , dts^2 + t^2 + 2(s+t) + 2$$
Now switch to polar coordinates $s = r cos theta, , t = r sin theta $.
The sector $S = (r,theta): 0 leqslant r leqslant J, ,0 leqslant theta leqslant pi/2$ is a subset of the square $[0,J]^2$. Since the integrand is nonnegative and $1 leqslant cos theta + sin theta leqslant sqrt2$ for $0 leqslant theta leqslant pi/2$ we have
$$beginalignA_I,J &geqslant int_0^pi/2int_0^J fracr, dr, dthetar^2 + 2r(cos theta + sin theta)+2\ &geqslant int_0^pi/2int_0^J fracr, dr, dthetar^2 + 2sqrt2r+2 \ &= fracpi2int_0^Jfracr(r+sqrt2)^2 , dr \ &=fracpi2int_0^Jfrac1r+sqrt2 , dr - fracpi2int_0^Jfracsqrt2(r+sqrt2)^2 , dr \ &= fracpi2left(log(J + sqrt2) + fracsqrt2J + sqrt2 - (1 + log sqrt2) right)endalign$$
The RHS tends to $+infty$ as $I,J to infty$.
answered Aug 3 at 17:09
RRL
43.4k42260
Thank you. These are the details I was interested in seeing.
– user28763
Aug 3 at 17:22
add a comment |Â
up vote
1
down vote
accepted
Assuming WLOG that $J leqslant I$ and changing variables as $x = 1+s, , y = 1+t$, we have
$$beginalign A_I,J=int_1^I+1int_1^J+1 fracdx , dyx^2 + y^2endalign geqslant int_0^Jint_0^J fracds , dts^2 + t^2 + 2(s+t) + 2$$
Now switch to polar coordinates $s = r cos theta, , t = r sin theta $.
The sector $S = (r,theta): 0 leqslant r leqslant J, ,0 leqslant theta leqslant pi/2$ is a subset of the square $[0,J]^2$. Since the integrand is nonnegative and $1 leqslant cos theta + sin theta leqslant sqrt2$ for $0 leqslant theta leqslant pi/2$ we have
$$beginalignA_I,J &geqslant int_0^pi/2int_0^J fracr, dr, dthetar^2 + 2r(cos theta + sin theta)+2\ &geqslant int_0^pi/2int_0^J fracr, dr, dthetar^2 + 2sqrt2r+2 \ &= fracpi2int_0^Jfracr(r+sqrt2)^2 , dr \ &=fracpi2int_0^Jfrac1r+sqrt2 , dr - fracpi2int_0^Jfracsqrt2(r+sqrt2)^2 , dr \ &= fracpi2left(log(J + sqrt2) + fracsqrt2J + sqrt2 - (1 + log sqrt2) right)endalign$$
The RHS tends to $+infty$ as $I,J to infty$.
answered Aug 3 at 17:09
RRL
43.4k42260
Thank you. These are the details I was interested in seeing.
– user28763
Aug 3 at 17:22
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Assuming WLOG that $J leqslant I$ and changing variables as $x = 1+s, , y = 1+t$, we have
$$beginalign A_I,J=int_1^I+1int_1^J+1 fracdx , dyx^2 + y^2endalign geqslant int_0^Jint_0^J fracds , dts^2 + t^2 + 2(s+t) + 2$$
Now switch to polar coordinates $s = r cos theta, , t = r sin theta $.
The sector $S = (r,theta): 0 leqslant r leqslant J, ,0 leqslant theta leqslant pi/2$ is a subset of the square $[0,J]^2$. Since the integrand is nonnegative and $1 leqslant cos theta + sin theta leqslant sqrt2$ for $0 leqslant theta leqslant pi/2$ we have
$$beginalignA_I,J &geqslant int_0^pi/2int_0^J fracr, dr, dthetar^2 + 2r(cos theta + sin theta)+2\ &geqslant int_0^pi/2int_0^J fracr, dr, dthetar^2 + 2sqrt2r+2 \ &= fracpi2int_0^Jfracr(r+sqrt2)^2 , dr \ &=fracpi2int_0^Jfrac1r+sqrt2 , dr - fracpi2int_0^Jfracsqrt2(r+sqrt2)^2 , dr \ &= fracpi2left(log(J + sqrt2) + fracsqrt2J + sqrt2 - (1 + log sqrt2) right)endalign$$
The RHS tends to $+infty$ as $I,J to infty$.
answered Aug 3 at 17:09
RRL
43.4k42260
Assuming WLOG that $J leqslant I$ and changing variables as $x = 1+s, , y = 1+t$, we have
$$beginalign A_I,J=int_1^I+1int_1^J+1 fracdx , dyx^2 + y^2endalign geqslant int_0^Jint_0^J fracds , dts^2 + t^2 + 2(s+t) + 2$$
Now switch to polar coordinates $s = r cos theta, , t = r sin theta $.
The sector $S = (r,theta): 0 leqslant r leqslant J, ,0 leqslant theta leqslant pi/2$ is a subset of the square $[0,J]^2$. Since the integrand is nonnegative and $1 leqslant cos theta + sin theta leqslant sqrt2$ for $0 leqslant theta leqslant pi/2$ we have
$$beginalignA_I,J &geqslant int_0^pi/2int_0^J fracr, dr, dthetar^2 + 2r(cos theta + sin theta)+2\ &geqslant int_0^pi/2int_0^J fracr, dr, dthetar^2 + 2sqrt2r+2 \ &= fracpi2int_0^Jfracr(r+sqrt2)^2 , dr \ &=fracpi2int_0^Jfrac1r+sqrt2 , dr - fracpi2int_0^Jfracsqrt2(r+sqrt2)^2 , dr \ &= fracpi2left(log(J + sqrt2) + fracsqrt2J + sqrt2 - (1 + log sqrt2) right)endalign$$
The RHS tends to $+infty$ as $I,J to infty$.
answered Aug 3 at 17:09
RRL
43.4k42260
answered Aug 3 at 17:09
RRL
43.4k42260
answered Aug 3 at 17:09
RRL
43.4k42260
answered Aug 3 at 17:09
RRL
43.4k42260
43.4k42260
Thank you. These are the details I was interested in seeing.
– user28763
Aug 3 at 17:22
add a comment |Â
Thank you. These are the details I was interested in seeing.
– user28763
Aug 3 at 17:22
Thank you. These are the details I was interested in seeing.
– user28763
Aug 3 at 17:22
Thank you. These are the details I was interested in seeing.
– user28763
Aug 3 at 17:22
add a comment |Â
up vote
0
down vote
$int_1^I+1int_1^J+1fracdxdyx^2+y^2gt fracpi2int_sqrt2^min(I,J)+1fracdrr$ which becomes infinite like $log(min(I,J)+1)$ as both indicies become infinite.
answered Aug 2 at 0:31
herb steinberg
93529
add a comment |Â
up vote
0
down vote
$int_1^I+1int_1^J+1fracdxdyx^2+y^2gt fracpi2int_sqrt2^min(I,J)+1fracdrr$ which becomes infinite like $log(min(I,J)+1)$ as both indicies become infinite.
answered Aug 2 at 0:31
herb steinberg
93529
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$int_1^I+1int_1^J+1fracdxdyx^2+y^2gt fracpi2int_sqrt2^min(I,J)+1fracdrr$ which becomes infinite like $log(min(I,J)+1)$ as both indicies become infinite.
answered Aug 2 at 0:31
herb steinberg
93529
$int_1^I+1int_1^J+1fracdxdyx^2+y^2gt fracpi2int_sqrt2^min(I,J)+1fracdrr$ which becomes infinite like $log(min(I,J)+1)$ as both indicies become infinite.
answered Aug 2 at 0:31
herb steinberg
93529
edited Aug 2 at 9:04
answered Aug 2 at 0:31
herb steinberg
93529
answered Aug 2 at 0:31
herb steinberg
93529
answered Aug 2 at 0:31
herb steinberg
93529
93529
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2869588%2fdivergent-double-series-sum-i-1i-sum-j-1j-frac1i2-j2-int-1i%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
2
Out of curiosity, why not use the fact that $arctan'(x) = frac11+x^2$ instead of switching to polar coordinates?
– Clement C.
Aug 1 at 23:12
1
Hint: the limit as $I,Jto infty$ is the integral over $[1, infty) times [1, infty)$ - and that contains, for example, the region $fracpi6 le theta le fracpi3, r ge 2$.
– Daniel Schepler
Aug 2 at 0:47