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Evaluating the sum $sum_n=1^inftydfrac(-1)^nn^2$
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I am tasked to evaluate the sum
$$sum_n=1^inftydfrac(-1)^nn^2$$
Using contour integration. This is what I've done so far.
Let $C_N$ be the square defined by the lines $x=pm(N+tfrac12)pi$ and $y=pm(N+tfrac12)pi$. Let $f=tfrac1z^2sin(z)$. I was able to prove that
$$int_C_Ndfrac1z^2sin(z)dz=2pi ileft[dfrac16+2sum_n=1^Ndfrac(-1)^npi^2n^2 right]. $$
By Cauchy's residue theorem. So now if I could prove that he integral converges to zero then I would be done. The problem is that I can't seem to get an upper bound on the integral. I've having trouble with the fact that $|f|$ is unbounded within $C_N$ so ML inequality doesn't help me. Any hints would be great. Thanks.
complex-analysis contour-integration
edited Dec 28 '13 at 12:55
Zaid Alyafeai
9,18622166
asked Dec 28 '13 at 0:27
TheNumber23
1,878718
add a comment |Â
up vote
4
down vote
favorite
I am tasked to evaluate the sum
$$sum_n=1^inftydfrac(-1)^nn^2$$
Using contour integration. This is what I've done so far.
Let $C_N$ be the square defined by the lines $x=pm(N+tfrac12)pi$ and $y=pm(N+tfrac12)pi$. Let $f=tfrac1z^2sin(z)$. I was able to prove that
$$int_C_Ndfrac1z^2sin(z)dz=2pi ileft[dfrac16+2sum_n=1^Ndfrac(-1)^npi^2n^2 right]. $$
By Cauchy's residue theorem. So now if I could prove that he integral converges to zero then I would be done. The problem is that I can't seem to get an upper bound on the integral. I've having trouble with the fact that $|f|$ is unbounded within $C_N$ so ML inequality doesn't help me. Any hints would be great. Thanks.
complex-analysis contour-integration
edited Dec 28 '13 at 12:55
Zaid Alyafeai
9,18622166
asked Dec 28 '13 at 0:27
TheNumber23
1,878718
Was that function given to you or you came up with it?
– DonAntonio
Dec 28 '13 at 0:37
Are you sure the index starts at 0? Because the first term is in determinant.
– Ali Caglayan
Dec 28 '13 at 0:47
*indeterminate $$
– anon
Dec 28 '13 at 0:51
@DonAntonio I was given the function.
– TheNumber23
Dec 28 '13 at 1:07
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I am tasked to evaluate the sum
$$sum_n=1^inftydfrac(-1)^nn^2$$
Using contour integration. This is what I've done so far.
Let $C_N$ be the square defined by the lines $x=pm(N+tfrac12)pi$ and $y=pm(N+tfrac12)pi$. Let $f=tfrac1z^2sin(z)$. I was able to prove that
$$int_C_Ndfrac1z^2sin(z)dz=2pi ileft[dfrac16+2sum_n=1^Ndfrac(-1)^npi^2n^2 right]. $$
By Cauchy's residue theorem. So now if I could prove that he integral converges to zero then I would be done. The problem is that I can't seem to get an upper bound on the integral. I've having trouble with the fact that $|f|$ is unbounded within $C_N$ so ML inequality doesn't help me. Any hints would be great. Thanks.
complex-analysis contour-integration
edited Dec 28 '13 at 12:55
Zaid Alyafeai
9,18622166
asked Dec 28 '13 at 0:27
TheNumber23
1,878718
I am tasked to evaluate the sum
$$sum_n=1^inftydfrac(-1)^nn^2$$
Using contour integration. This is what I've done so far.
Let $C_N$ be the square defined by the lines $x=pm(N+tfrac12)pi$ and $y=pm(N+tfrac12)pi$. Let $f=tfrac1z^2sin(z)$. I was able to prove that
$$int_C_Ndfrac1z^2sin(z)dz=2pi ileft[dfrac16+2sum_n=1^Ndfrac(-1)^npi^2n^2 right]. $$
By Cauchy's residue theorem. So now if I could prove that he integral converges to zero then I would be done. The problem is that I can't seem to get an upper bound on the integral. I've having trouble with the fact that $|f|$ is unbounded within $C_N$ so ML inequality doesn't help me. Any hints would be great. Thanks.
complex-analysis contour-integration
edited Dec 28 '13 at 12:55
Zaid Alyafeai
9,18622166
asked Dec 28 '13 at 0:27
TheNumber23
1,878718
edited Dec 28 '13 at 12:55
Zaid Alyafeai
9,18622166
edited Dec 28 '13 at 12:55
Zaid Alyafeai
9,18622166
edited Dec 28 '13 at 12:55
Zaid Alyafeai
9,18622166
9,18622166
asked Dec 28 '13 at 0:27
TheNumber23
1,878718
asked Dec 28 '13 at 0:27
TheNumber23
1,878718
asked Dec 28 '13 at 0:27
TheNumber23
1,878718
1,878718
Was that function given to you or you came up with it?
– DonAntonio
Dec 28 '13 at 0:37
Are you sure the index starts at 0? Because the first term is in determinant.
– Ali Caglayan
Dec 28 '13 at 0:47
*indeterminate $$
– anon
Dec 28 '13 at 0:51
@DonAntonio I was given the function.
– TheNumber23
Dec 28 '13 at 1:07
add a comment |Â
Was that function given to you or you came up with it?
– DonAntonio
Dec 28 '13 at 0:37
Are you sure the index starts at 0? Because the first term is in determinant.
– Ali Caglayan
Dec 28 '13 at 0:47
*indeterminate $$
– anon
Dec 28 '13 at 0:51
@DonAntonio I was given the function.
– TheNumber23
Dec 28 '13 at 1:07
Was that function given to you or you came up with it?
– DonAntonio
Dec 28 '13 at 0:37
Was that function given to you or you came up with it?
– DonAntonio
Dec 28 '13 at 0:37
Are you sure the index starts at 0? Because the first term is in determinant.
– Ali Caglayan
Dec 28 '13 at 0:47
Are you sure the index starts at 0? Because the first term is in determinant.
– Ali Caglayan
Dec 28 '13 at 0:47
*indeterminate $$
– anon
Dec 28 '13 at 0:51
*indeterminate $$
– anon
Dec 28 '13 at 0:51
@DonAntonio I was given the function.
– TheNumber23
Dec 28 '13 at 1:07
@DonAntonio I was given the function.
– TheNumber23
Dec 28 '13 at 1:07
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
On the contour $C_N$, we have $lvert z^2rvert geqslant left(N+frac12right)^2pi^2$, so if we can bound the modulus of the sine factor below, say $lvert sin z rvert geqslant varepsilon > 0$ on $C_N$, the standard estimate yields
$$leftlvert int_C_N fracdzz^2sin zrightrvert leqslant frac4cdot(2N+1)pileft(N+frac12right)^2pi^2varepsilon = frac8left(N+frac12right)pivarepsilon xrightarrowNtoinfty 0.$$
But with $z = x+iy$, we have $sin z = sin x cos (iy) + cos xsin (iy) = sin x cosh y + icos xsinh y$, so
$$lvert sin zrvert^2 = sin^2 x cosh^2 y + cos^2 x sinh^2y = sin^2 x + sinh^2 y.$$
Now, on the vertical sides of the contour $C_N$ ($x = pm left(N+frac12right)pi$) we have $sin x = pm 1$, so $lvert sin zrvert geqslant 1$ on those. On the horizontal sides, we have $lvert sin zrvert geqslant sinh left(left(N+frac12right)piright) geqslant sinh fracpi2 > 2$, so we can choose $varepsilon = 1$ in the above.
answered Dec 28 '13 at 13:15
Daniel Fischer♦
171k16154274
add a comment |Â
up vote
0
down vote
$Res(frac 1z^2sin z,0)=0$. Write it like this $$f(z)=frac 1z^2sin z=frac 1z^2(z-frac z^33+frac z^55-...)=frac 1z^3(1-frac z^23+frac z^45-...)$$. So $f$ has a pole of order 3 at $0$.
answered Dec 28 '13 at 0:50
Haha
4,8071015
I've already proved the integral result. And calculated the residue at 0.
– TheNumber23
Dec 28 '13 at 1:08
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
On the contour $C_N$, we have $lvert z^2rvert geqslant left(N+frac12right)^2pi^2$, so if we can bound the modulus of the sine factor below, say $lvert sin z rvert geqslant varepsilon > 0$ on $C_N$, the standard estimate yields
$$leftlvert int_C_N fracdzz^2sin zrightrvert leqslant frac4cdot(2N+1)pileft(N+frac12right)^2pi^2varepsilon = frac8left(N+frac12right)pivarepsilon xrightarrowNtoinfty 0.$$
But with $z = x+iy$, we have $sin z = sin x cos (iy) + cos xsin (iy) = sin x cosh y + icos xsinh y$, so
$$lvert sin zrvert^2 = sin^2 x cosh^2 y + cos^2 x sinh^2y = sin^2 x + sinh^2 y.$$
Now, on the vertical sides of the contour $C_N$ ($x = pm left(N+frac12right)pi$) we have $sin x = pm 1$, so $lvert sin zrvert geqslant 1$ on those. On the horizontal sides, we have $lvert sin zrvert geqslant sinh left(left(N+frac12right)piright) geqslant sinh fracpi2 > 2$, so we can choose $varepsilon = 1$ in the above.
answered Dec 28 '13 at 13:15
Daniel Fischer♦
171k16154274
add a comment |Â
up vote
2
down vote
accepted
On the contour $C_N$, we have $lvert z^2rvert geqslant left(N+frac12right)^2pi^2$, so if we can bound the modulus of the sine factor below, say $lvert sin z rvert geqslant varepsilon > 0$ on $C_N$, the standard estimate yields
$$leftlvert int_C_N fracdzz^2sin zrightrvert leqslant frac4cdot(2N+1)pileft(N+frac12right)^2pi^2varepsilon = frac8left(N+frac12right)pivarepsilon xrightarrowNtoinfty 0.$$
But with $z = x+iy$, we have $sin z = sin x cos (iy) + cos xsin (iy) = sin x cosh y + icos xsinh y$, so
$$lvert sin zrvert^2 = sin^2 x cosh^2 y + cos^2 x sinh^2y = sin^2 x + sinh^2 y.$$
Now, on the vertical sides of the contour $C_N$ ($x = pm left(N+frac12right)pi$) we have $sin x = pm 1$, so $lvert sin zrvert geqslant 1$ on those. On the horizontal sides, we have $lvert sin zrvert geqslant sinh left(left(N+frac12right)piright) geqslant sinh fracpi2 > 2$, so we can choose $varepsilon = 1$ in the above.
answered Dec 28 '13 at 13:15
Daniel Fischer♦
171k16154274
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
On the contour $C_N$, we have $lvert z^2rvert geqslant left(N+frac12right)^2pi^2$, so if we can bound the modulus of the sine factor below, say $lvert sin z rvert geqslant varepsilon > 0$ on $C_N$, the standard estimate yields
$$leftlvert int_C_N fracdzz^2sin zrightrvert leqslant frac4cdot(2N+1)pileft(N+frac12right)^2pi^2varepsilon = frac8left(N+frac12right)pivarepsilon xrightarrowNtoinfty 0.$$
But with $z = x+iy$, we have $sin z = sin x cos (iy) + cos xsin (iy) = sin x cosh y + icos xsinh y$, so
$$lvert sin zrvert^2 = sin^2 x cosh^2 y + cos^2 x sinh^2y = sin^2 x + sinh^2 y.$$
Now, on the vertical sides of the contour $C_N$ ($x = pm left(N+frac12right)pi$) we have $sin x = pm 1$, so $lvert sin zrvert geqslant 1$ on those. On the horizontal sides, we have $lvert sin zrvert geqslant sinh left(left(N+frac12right)piright) geqslant sinh fracpi2 > 2$, so we can choose $varepsilon = 1$ in the above.
answered Dec 28 '13 at 13:15
Daniel Fischer♦
171k16154274
On the contour $C_N$, we have $lvert z^2rvert geqslant left(N+frac12right)^2pi^2$, so if we can bound the modulus of the sine factor below, say $lvert sin z rvert geqslant varepsilon > 0$ on $C_N$, the standard estimate yields
$$leftlvert int_C_N fracdzz^2sin zrightrvert leqslant frac4cdot(2N+1)pileft(N+frac12right)^2pi^2varepsilon = frac8left(N+frac12right)pivarepsilon xrightarrowNtoinfty 0.$$
But with $z = x+iy$, we have $sin z = sin x cos (iy) + cos xsin (iy) = sin x cosh y + icos xsinh y$, so
$$lvert sin zrvert^2 = sin^2 x cosh^2 y + cos^2 x sinh^2y = sin^2 x + sinh^2 y.$$
Now, on the vertical sides of the contour $C_N$ ($x = pm left(N+frac12right)pi$) we have $sin x = pm 1$, so $lvert sin zrvert geqslant 1$ on those. On the horizontal sides, we have $lvert sin zrvert geqslant sinh left(left(N+frac12right)piright) geqslant sinh fracpi2 > 2$, so we can choose $varepsilon = 1$ in the above.
answered Dec 28 '13 at 13:15
Daniel Fischer♦
171k16154274
answered Dec 28 '13 at 13:15
Daniel Fischer♦
171k16154274
answered Dec 28 '13 at 13:15
Daniel Fischer♦
171k16154274
answered Dec 28 '13 at 13:15
Daniel Fischer♦
171k16154274
171k16154274
add a comment |Â
add a comment |Â
up vote
0
down vote
$Res(frac 1z^2sin z,0)=0$. Write it like this $$f(z)=frac 1z^2sin z=frac 1z^2(z-frac z^33+frac z^55-...)=frac 1z^3(1-frac z^23+frac z^45-...)$$. So $f$ has a pole of order 3 at $0$.
answered Dec 28 '13 at 0:50
Haha
4,8071015
I've already proved the integral result. And calculated the residue at 0.
– TheNumber23
Dec 28 '13 at 1:08
add a comment |Â
up vote
0
down vote
$Res(frac 1z^2sin z,0)=0$. Write it like this $$f(z)=frac 1z^2sin z=frac 1z^2(z-frac z^33+frac z^55-...)=frac 1z^3(1-frac z^23+frac z^45-...)$$. So $f$ has a pole of order 3 at $0$.
answered Dec 28 '13 at 0:50
Haha
4,8071015
I've already proved the integral result. And calculated the residue at 0.
– TheNumber23
Dec 28 '13 at 1:08
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$Res(frac 1z^2sin z,0)=0$. Write it like this $$f(z)=frac 1z^2sin z=frac 1z^2(z-frac z^33+frac z^55-...)=frac 1z^3(1-frac z^23+frac z^45-...)$$. So $f$ has a pole of order 3 at $0$.
answered Dec 28 '13 at 0:50
Haha
4,8071015
$Res(frac 1z^2sin z,0)=0$. Write it like this $$f(z)=frac 1z^2sin z=frac 1z^2(z-frac z^33+frac z^55-...)=frac 1z^3(1-frac z^23+frac z^45-...)$$. So $f$ has a pole of order 3 at $0$.
answered Dec 28 '13 at 0:50
Haha
4,8071015
answered Dec 28 '13 at 0:50
Haha
4,8071015
answered Dec 28 '13 at 0:50
Haha
4,8071015
answered Dec 28 '13 at 0:50
Haha
4,8071015
4,8071015
I've already proved the integral result. And calculated the residue at 0.
– TheNumber23
Dec 28 '13 at 1:08
add a comment |Â
I've already proved the integral result. And calculated the residue at 0.
– TheNumber23
Dec 28 '13 at 1:08
I've already proved the integral result. And calculated the residue at 0.
– TheNumber23
Dec 28 '13 at 1:08
I've already proved the integral result. And calculated the residue at 0.
– TheNumber23
Dec 28 '13 at 1:08
add a comment |Â
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Was that function given to you or you came up with it?
– DonAntonio
Dec 28 '13 at 0:37
Are you sure the index starts at 0? Because the first term is in determinant.
– Ali Caglayan
Dec 28 '13 at 0:47
*indeterminate $$
– anon
Dec 28 '13 at 0:51
@DonAntonio I was given the function.
– TheNumber23
Dec 28 '13 at 1:07