Mixing
Use Residue Theorem to Sum Series
$begingroup$
Show that
$$sum_{n=- infty}^{infty} frac{1}{(3n-1)^2} = frac{4 pi^2}{27}$$
I'm pretty sure I need to use the Residue Theorem to sum the series, but I'm unsure how to apply it. Here's what I know:
- There is a double pole at $3n-1=0 Rightarrow n= frac{1}{3}$
- $sum_{n=-infty}^{infty} f(n) = -sum$ of the resides of $ pi cot( pi z) cdot f(z)$ at the pole(s) of $f(z)$
Using the above information, here's what I've done to calculate the residue at $z= frac{1}{3}$ so far:
$$ frac{1}{(2-1)!}lim_{ztofrac{1}{3}} Bigg[ frac{d^{2-1}}{dz^{2-1}} Bigg((z- frac{1}{3}) pi cot( pi z) frac{1}{{(3z-1)}^2} Bigg) Bigg]$$
$$= Bigg(lim_{ztofrac{1}{3}} frac{d}{dz} pi cot (pi z)Bigg) Bigg(lim_{ztofrac{1}{3}} frac{d}{dz} frac{z-frac{1}{3}}{(3z-1)^2}Bigg) $$
$$= Bigg( lim_{ztofrac{1}{3}} {- pi^2 csc({pi z})} Bigg) Bigg( lim_{ztofrac{1}{3}} frac{1}{(3z-1)^2}Bigg)$$
But
$$lim_{ztofrac{1}{3}} frac{1}{(3z-1)^2}= infty$$
I know I've gone wrong somewhere (maybe when I split it up into two limits?), but I'm not sure where. Any help or guidance would be greatly appreciated. Thank you!!
sequences-and-series complex-analysis residue-calculus
asked May 18 '16 at 19:08
user26535user26535
212
$endgroup$
add a comment |
$begingroup$
Show that
$$sum_{n=- infty}^{infty} frac{1}{(3n-1)^2} = frac{4 pi^2}{27}$$
I'm pretty sure I need to use the Residue Theorem to sum the series, but I'm unsure how to apply it. Here's what I know:
- There is a double pole at $3n-1=0 Rightarrow n= frac{1}{3}$
- $sum_{n=-infty}^{infty} f(n) = -sum$ of the resides of $ pi cot( pi z) cdot f(z)$ at the pole(s) of $f(z)$
Using the above information, here's what I've done to calculate the residue at $z= frac{1}{3}$ so far:
$$ frac{1}{(2-1)!}lim_{ztofrac{1}{3}} Bigg[ frac{d^{2-1}}{dz^{2-1}} Bigg((z- frac{1}{3}) pi cot( pi z) frac{1}{{(3z-1)}^2} Bigg) Bigg]$$
$$= Bigg(lim_{ztofrac{1}{3}} frac{d}{dz} pi cot (pi z)Bigg) Bigg(lim_{ztofrac{1}{3}} frac{d}{dz} frac{z-frac{1}{3}}{(3z-1)^2}Bigg) $$
$$= Bigg( lim_{ztofrac{1}{3}} {- pi^2 csc({pi z})} Bigg) Bigg( lim_{ztofrac{1}{3}} frac{1}{(3z-1)^2}Bigg)$$
But
$$lim_{ztofrac{1}{3}} frac{1}{(3z-1)^2}= infty$$
I know I've gone wrong somewhere (maybe when I split it up into two limits?), but I'm not sure where. Any help or guidance would be greatly appreciated. Thank you!!
sequences-and-series complex-analysis residue-calculus
asked May 18 '16 at 19:08
user26535user26535
212
$endgroup$
$begingroup$
At a pole $z_0$ of $g$ of order $k > 0$, you need $$biggl(frac{d}{dz}biggr)^{k-1}bigl( (z - z_0)^kcdot g(z)bigr).$$ You used $(z - z_0)^1$ instead.
$endgroup$
– Daniel Fischer♦
May 18 '16 at 19:12
1
$begingroup$
But, for this series, you can have it easier. For $n > 1$, you get $frac{1}{2^2},, frac{1}{5^2},, frac{1}{8^2},,dotsc$, and for $n leqslant 0$ you get $frac{1}{(-1)^2},, frac{1}{(-4)^2},, frac{1}{(-7)^2},,dotsc$. Since $(-(3k+1))^2 = (3k+1)^2$, if you add, what's the difference to $sum_{m = 1}^{infty} frac{1}{m^2} = zeta(2)$?
$endgroup$
– Daniel Fischer♦
May 18 '16 at 19:15
$begingroup$
@DanielFischer Got it. Silly mistake. Thanks for your help!
$endgroup$
– user26535
May 18 '16 at 19:28
$begingroup$
This question was already asked earlier yesterday o the day before. You don't need complex theory to solve this, yet you need, at least with my method, to know that $$sum_{n=1}^inftyfrac1{n^2}=frac{pi^2}6$$
$endgroup$
– DonAntonio
May 19 '16 at 21:45
add a comment |
$begingroup$
Show that
$$sum_{n=- infty}^{infty} frac{1}{(3n-1)^2} = frac{4 pi^2}{27}$$
I'm pretty sure I need to use the Residue Theorem to sum the series, but I'm unsure how to apply it. Here's what I know:
- There is a double pole at $3n-1=0 Rightarrow n= frac{1}{3}$
- $sum_{n=-infty}^{infty} f(n) = -sum$ of the resides of $ pi cot( pi z) cdot f(z)$ at the pole(s) of $f(z)$
Using the above information, here's what I've done to calculate the residue at $z= frac{1}{3}$ so far:
$$ frac{1}{(2-1)!}lim_{ztofrac{1}{3}} Bigg[ frac{d^{2-1}}{dz^{2-1}} Bigg((z- frac{1}{3}) pi cot( pi z) frac{1}{{(3z-1)}^2} Bigg) Bigg]$$
$$= Bigg(lim_{ztofrac{1}{3}} frac{d}{dz} pi cot (pi z)Bigg) Bigg(lim_{ztofrac{1}{3}} frac{d}{dz} frac{z-frac{1}{3}}{(3z-1)^2}Bigg) $$
$$= Bigg( lim_{ztofrac{1}{3}} {- pi^2 csc({pi z})} Bigg) Bigg( lim_{ztofrac{1}{3}} frac{1}{(3z-1)^2}Bigg)$$
But
$$lim_{ztofrac{1}{3}} frac{1}{(3z-1)^2}= infty$$
I know I've gone wrong somewhere (maybe when I split it up into two limits?), but I'm not sure where. Any help or guidance would be greatly appreciated. Thank you!!
sequences-and-series complex-analysis residue-calculus
asked May 18 '16 at 19:08
user26535user26535
212
$endgroup$
Show that
$$sum_{n=- infty}^{infty} frac{1}{(3n-1)^2} = frac{4 pi^2}{27}$$
I'm pretty sure I need to use the Residue Theorem to sum the series, but I'm unsure how to apply it. Here's what I know:
- There is a double pole at $3n-1=0 Rightarrow n= frac{1}{3}$
- $sum_{n=-infty}^{infty} f(n) = -sum$ of the resides of $ pi cot( pi z) cdot f(z)$ at the pole(s) of $f(z)$
Using the above information, here's what I've done to calculate the residue at $z= frac{1}{3}$ so far:
$$ frac{1}{(2-1)!}lim_{ztofrac{1}{3}} Bigg[ frac{d^{2-1}}{dz^{2-1}} Bigg((z- frac{1}{3}) pi cot( pi z) frac{1}{{(3z-1)}^2} Bigg) Bigg]$$
$$= Bigg(lim_{ztofrac{1}{3}} frac{d}{dz} pi cot (pi z)Bigg) Bigg(lim_{ztofrac{1}{3}} frac{d}{dz} frac{z-frac{1}{3}}{(3z-1)^2}Bigg) $$
$$= Bigg( lim_{ztofrac{1}{3}} {- pi^2 csc({pi z})} Bigg) Bigg( lim_{ztofrac{1}{3}} frac{1}{(3z-1)^2}Bigg)$$
But
$$lim_{ztofrac{1}{3}} frac{1}{(3z-1)^2}= infty$$
I know I've gone wrong somewhere (maybe when I split it up into two limits?), but I'm not sure where. Any help or guidance would be greatly appreciated. Thank you!!
sequences-and-series complex-analysis residue-calculus
sequences-and-series complex-analysis residue-calculus
asked May 18 '16 at 19:08
user26535user26535
212
asked May 18 '16 at 19:08
user26535user26535
212
asked May 18 '16 at 19:08
user26535user26535
212
asked May 18 '16 at 19:08
user26535user26535
212
asked May 18 '16 at 19:08
user26535user26535
212
212
$begingroup$
At a pole $z_0$ of $g$ of order $k > 0$, you need $$biggl(frac{d}{dz}biggr)^{k-1}bigl( (z - z_0)^kcdot g(z)bigr).$$ You used $(z - z_0)^1$ instead.
$endgroup$
– Daniel Fischer♦
May 18 '16 at 19:12
1
$begingroup$
But, for this series, you can have it easier. For $n > 1$, you get $frac{1}{2^2},, frac{1}{5^2},, frac{1}{8^2},,dotsc$, and for $n leqslant 0$ you get $frac{1}{(-1)^2},, frac{1}{(-4)^2},, frac{1}{(-7)^2},,dotsc$. Since $(-(3k+1))^2 = (3k+1)^2$, if you add, what's the difference to $sum_{m = 1}^{infty} frac{1}{m^2} = zeta(2)$?
$endgroup$
– Daniel Fischer♦
May 18 '16 at 19:15
$begingroup$
@DanielFischer Got it. Silly mistake. Thanks for your help!
$endgroup$
– user26535
May 18 '16 at 19:28
$begingroup$
This question was already asked earlier yesterday o the day before. You don't need complex theory to solve this, yet you need, at least with my method, to know that $$sum_{n=1}^inftyfrac1{n^2}=frac{pi^2}6$$
$endgroup$
– DonAntonio
May 19 '16 at 21:45
add a comment |
$begingroup$
At a pole $z_0$ of $g$ of order $k > 0$, you need $$biggl(frac{d}{dz}biggr)^{k-1}bigl( (z - z_0)^kcdot g(z)bigr).$$ You used $(z - z_0)^1$ instead.
$endgroup$
– Daniel Fischer♦
May 18 '16 at 19:12
1
$begingroup$
But, for this series, you can have it easier. For $n > 1$, you get $frac{1}{2^2},, frac{1}{5^2},, frac{1}{8^2},,dotsc$, and for $n leqslant 0$ you get $frac{1}{(-1)^2},, frac{1}{(-4)^2},, frac{1}{(-7)^2},,dotsc$. Since $(-(3k+1))^2 = (3k+1)^2$, if you add, what's the difference to $sum_{m = 1}^{infty} frac{1}{m^2} = zeta(2)$?
$endgroup$
– Daniel Fischer♦
May 18 '16 at 19:15
$begingroup$
@DanielFischer Got it. Silly mistake. Thanks for your help!
$endgroup$
– user26535
May 18 '16 at 19:28
$begingroup$
This question was already asked earlier yesterday o the day before. You don't need complex theory to solve this, yet you need, at least with my method, to know that $$sum_{n=1}^inftyfrac1{n^2}=frac{pi^2}6$$
$endgroup$
– DonAntonio
May 19 '16 at 21:45
$begingroup$
At a pole $z_0$ of $g$ of order $k > 0$, you need $$biggl(frac{d}{dz}biggr)^{k-1}bigl( (z - z_0)^kcdot g(z)bigr).$$ You used $(z - z_0)^1$ instead.
$endgroup$
– Daniel Fischer♦
May 18 '16 at 19:12
$begingroup$
At a pole $z_0$ of $g$ of order $k > 0$, you need $$biggl(frac{d}{dz}biggr)^{k-1}bigl( (z - z_0)^kcdot g(z)bigr).$$ You used $(z - z_0)^1$ instead.
$endgroup$
– Daniel Fischer♦
May 18 '16 at 19:12
1
1
$begingroup$
But, for this series, you can have it easier. For $n > 1$, you get $frac{1}{2^2},, frac{1}{5^2},, frac{1}{8^2},,dotsc$, and for $n leqslant 0$ you get $frac{1}{(-1)^2},, frac{1}{(-4)^2},, frac{1}{(-7)^2},,dotsc$. Since $(-(3k+1))^2 = (3k+1)^2$, if you add, what's the difference to $sum_{m = 1}^{infty} frac{1}{m^2} = zeta(2)$?
$endgroup$
– Daniel Fischer♦
May 18 '16 at 19:15
$begingroup$
But, for this series, you can have it easier. For $n > 1$, you get $frac{1}{2^2},, frac{1}{5^2},, frac{1}{8^2},,dotsc$, and for $n leqslant 0$ you get $frac{1}{(-1)^2},, frac{1}{(-4)^2},, frac{1}{(-7)^2},,dotsc$. Since $(-(3k+1))^2 = (3k+1)^2$, if you add, what's the difference to $sum_{m = 1}^{infty} frac{1}{m^2} = zeta(2)$?
$endgroup$
– Daniel Fischer♦
May 18 '16 at 19:15
$begingroup$
@DanielFischer Got it. Silly mistake. Thanks for your help!
$endgroup$
– user26535
May 18 '16 at 19:28
$begingroup$
@DanielFischer Got it. Silly mistake. Thanks for your help!
$endgroup$
– user26535
May 18 '16 at 19:28
$begingroup$
This question was already asked earlier yesterday o the day before. You don't need complex theory to solve this, yet you need, at least with my method, to know that $$sum_{n=1}^inftyfrac1{n^2}=frac{pi^2}6$$
$endgroup$
– DonAntonio
May 19 '16 at 21:45
$begingroup$
This question was already asked earlier yesterday o the day before. You don't need complex theory to solve this, yet you need, at least with my method, to know that $$sum_{n=1}^inftyfrac1{n^2}=frac{pi^2}6$$
$endgroup$
– DonAntonio
May 19 '16 at 21:45
add a comment |
1 Answer
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$begingroup$
Your sum $$sum_{n=-infty}^{infty}frac{1}{(3n-1)^2}$$ is no different from $$sum_{n=1}^{infty}frac{1}{n^2}-sum_{n=1}^{infty}frac{1}{(3n)^2}.$$ Now using user26535 $zeta(2)=frac{pi^2}{6}$ we can gather that your sum is the same as $(1-frac{1}{9})frac{pi^2}{6}$.
answered Oct 20 '17 at 10:51
Boon OngBoon Ong
11
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$begingroup$
Your sum $$sum_{n=-infty}^{infty}frac{1}{(3n-1)^2}$$ is no different from $$sum_{n=1}^{infty}frac{1}{n^2}-sum_{n=1}^{infty}frac{1}{(3n)^2}.$$ Now using user26535 $zeta(2)=frac{pi^2}{6}$ we can gather that your sum is the same as $(1-frac{1}{9})frac{pi^2}{6}$.
answered Oct 20 '17 at 10:51
Boon OngBoon Ong
11
$endgroup$
add a comment |
$begingroup$
Your sum $$sum_{n=-infty}^{infty}frac{1}{(3n-1)^2}$$ is no different from $$sum_{n=1}^{infty}frac{1}{n^2}-sum_{n=1}^{infty}frac{1}{(3n)^2}.$$ Now using user26535 $zeta(2)=frac{pi^2}{6}$ we can gather that your sum is the same as $(1-frac{1}{9})frac{pi^2}{6}$.
answered Oct 20 '17 at 10:51
Boon OngBoon Ong
11
$endgroup$
add a comment |
$begingroup$
Your sum $$sum_{n=-infty}^{infty}frac{1}{(3n-1)^2}$$ is no different from $$sum_{n=1}^{infty}frac{1}{n^2}-sum_{n=1}^{infty}frac{1}{(3n)^2}.$$ Now using user26535 $zeta(2)=frac{pi^2}{6}$ we can gather that your sum is the same as $(1-frac{1}{9})frac{pi^2}{6}$.
answered Oct 20 '17 at 10:51
Boon OngBoon Ong
11
$endgroup$
Your sum $$sum_{n=-infty}^{infty}frac{1}{(3n-1)^2}$$ is no different from $$sum_{n=1}^{infty}frac{1}{n^2}-sum_{n=1}^{infty}frac{1}{(3n)^2}.$$ Now using user26535 $zeta(2)=frac{pi^2}{6}$ we can gather that your sum is the same as $(1-frac{1}{9})frac{pi^2}{6}$.
answered Oct 20 '17 at 10:51
Boon OngBoon Ong
11
answered Oct 20 '17 at 10:51
Boon OngBoon Ong
11
answered Oct 20 '17 at 10:51
Boon OngBoon Ong
11
answered Oct 20 '17 at 10:51
Boon OngBoon Ong
11
11
add a comment |
add a comment |
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$begingroup$
At a pole $z_0$ of $g$ of order $k > 0$, you need $$biggl(frac{d}{dz}biggr)^{k-1}bigl( (z - z_0)^kcdot g(z)bigr).$$ You used $(z - z_0)^1$ instead.
$endgroup$
– Daniel Fischer♦
May 18 '16 at 19:12
1
$begingroup$
But, for this series, you can have it easier. For $n > 1$, you get $frac{1}{2^2},, frac{1}{5^2},, frac{1}{8^2},,dotsc$, and for $n leqslant 0$ you get $frac{1}{(-1)^2},, frac{1}{(-4)^2},, frac{1}{(-7)^2},,dotsc$. Since $(-(3k+1))^2 = (3k+1)^2$, if you add, what's the difference to $sum_{m = 1}^{infty} frac{1}{m^2} = zeta(2)$?
$endgroup$
– Daniel Fischer♦
May 18 '16 at 19:15
$begingroup$
@DanielFischer Got it. Silly mistake. Thanks for your help!
$endgroup$
– user26535
May 18 '16 at 19:28
$begingroup$
This question was already asked earlier yesterday o the day before. You don't need complex theory to solve this, yet you need, at least with my method, to know that $$sum_{n=1}^inftyfrac1{n^2}=frac{pi^2}6$$
$endgroup$
– DonAntonio
May 19 '16 at 21:45