It can be shown that n = 1 ∑ ∞ n 2 coth 2 π n = Q P G + S R π T where G = n = 0 ∑ ∞ ( 2 n + 1 ) 2 ( − 1 ) n = 0 . 9 1 5 9 6 5 5 9 4 … is Catalan's constant , and where P , Q , R , S and T are all positive integers with P , Q and R , S coprime pairs. Find P + Q + R + S + T .
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Well done! If the function can be differentiated as a function of z , except at some at most countable collection of isolated points, where the function takes the value ∞ , it is meromorphic. Meromorphic equals entire with some isolated singularities.
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Ohh I got it, Thanks !
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Stick with my final sentence: meromorphic equals entire except at some isolated singularities. The value of a function at an isolated essential singularity (infinite pole) is not ∞ ; the function takes values arbitrarily close to any complex number arbitrarily close to any essential singularity. Only at (finite) poles does the function tend to ∞ near the singularity.
It's wonderful, I used the same steps but I'm confused that mathematically how to determine the function is meromorphic or not so that we can apply residue theorem
I do not know contour yet, so I do not know how to generalise this sum.
Generalising this sum can give interesting results regarding the generating function of the divisor function. For instance:
n = 1 ∑ ∞ n 2 coth 2 ( 2 ln x n ) = 6 π 2 + 4 [ n = 1 ∑ ∞ n 2 σ 3 ( n ) x − n ]
n = 1 ∑ ∞ n 2 coth ( 2 ln x n ) = 6 π 2 + 2 p = 1 ∑ ∞ p 2 σ 2 ( p ) x − p
Consider
∫ C f ( z ) d z = ∫ C z 2 π cot ( π z ) coth 2 ( π z ) d z
Poles
At z = 0 order 5
At z = n , n ∈ Z − { 0 } order 1
At z = k i , k ∈ Z − { 0 } order 2
Evaluation
At z = 0 , Laurent series of f ( z ) = π 2 z 5 1 + 3 z 3 1 − 4 5 z 8 π 2 + ⋯
Residue is − 4 5 8 π 2
At z = n
z → n lim z 2 ( z − n ) π cot ( π z ) coth 2 ( π z ) = n 2 coth 2 ( π z )
At z = k i
z → k i lim d z d ( z 2 ( z − k i ) 2 π cot ( π z ) coth 2 ( π z ) ) = − π k 3 ( k π + sinh ( 2 k π ) ) \cosech 2 ( k π )
∫ C z 2 π cot ( π z ) coth 2 ( π z ) d z = 0 ∫ C z 2 π cot ( π z ) coth 2 ( π z ) d z = − 4 5 8 π 2 + 2 n ≥ 1 ∑ n 2 coth 2 ( π n ) − 2 k ≥ 1 ∑ π k 3 ( k π + sinh ( 2 k π ) ) \cosech 2 ( k π )
But k ≥ 1 ∑ π k 3 ( k π + sinh ( 2 k π ) ) \cosech 2 ( k π ) = 3 2 G + 1 0 1 ζ ( 2 ) = 3 2 G + 6 0 π 2
Thus, n ≥ 1 ∑ n 2 coth 2 ( π n ) = 4 5 4 π 2 + 3 2 G + 6 0 π 2
n ≥ 1 ∑ n 2 coth 2 ( π n ) = 3 2 G + 1 8 0 1 9 π 2
You just state a formula for k ≥ 1 ∑ π k 3 ( k π + sinh 2 k π ) c o s e c h 2 k π Where does that come from? If you combine it with the result for k ≥ 1 ∑ k 3 coth k π That I gave, and the identity c o s e c h 2 x = coth 2 x − 1 , then the result is immediate...
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You mean , you want proof for the formula ?
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Yes. If you have a direct proof of it, it might yield a much simpler derivation of the result, putting your and my approaches together. As it stands, it is a rather strong statement to be given without proof or reference.
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Let C N be the positively oriented square contour with vertices at ξ ( N + 2 1 ) + i η ( N + 2 1 ) for ξ , η = 1 , − 1 .
For any b ∈ N the function f b ( z ) = ( z 2 + b 2 ) 2 π cot π z is meromorphic with simple poles at every integer, and with double poles at ± i b . Thus 2 π i 1 ∫ C N f b ( z ) d z = = R e s z = i b f b ( z ) + R e s z = − i b f b ( z ) + a = − N ∑ N R e s z = a f b ( z ) 2 b 2 π 2 − 2 b 3 π coth π b − 2 b 2 π 2 coth 2 π b + b 4 1 + 2 a = 1 ∑ N ( a 2 + b 2 ) 2 1 Since ∣ f b ( z ) ∣ = O ( N − 4 ) uniformly on C N , it follows, letting N → ∞ , that a = 1 ∑ ∞ ( a 2 + b 2 ) 2 1 = 4 b 3 π coth π b + 4 b 2 π 2 coth 2 π b − 2 b 4 1 − 4 b 2 π 2 for any b ∈ N , and hence it follows that a , b = 1 ∑ ∞ ( a 2 + b 2 ) 2 1 = 4 1 π b = 1 ∑ ∞ b 3 coth π b + 4 1 π 2 b = 1 ∑ ∞ b 2 coth 2 π b − 2 1 ζ ( 4 ) − 4 1 π 2 ζ ( 2 )
Next, the function g ( z ) = z 3 π cot π z coth π z is meromorphic with simple poles at n and i n for any nonzero integer n , and a quintuple pole at 0 . Thus 2 π i 1 ∫ C N g ( z ) d z = = R e s z = 0 g ( z ) + b = 1 ∑ N ( R e s z = b + R e s z = − b + R e s z = i b + R e s z = − i b ) g ( z ) 4 b = 1 ∑ N b 3 coth π b − 4 5 7 π 3 Since ∣ g ( z ) ∣ = O ( N − 3 ) uniformly on C N , letting N → ∞ gives us that b = 1 ∑ ∞ b 3 coth π b = 1 8 0 7 π 3 and hence a , b = 1 ∑ ∞ ( a 2 + b 2 ) 2 1 = 4 1 π 2 b = 1 ∑ ∞ b 2 coth 2 π b − 8 0 3 π 4
Finally, using results from Number Theory we know that a , b = 1 ∑ ∞ ( a 2 + b 2 ) 2 1 = ζ ( 2 ) G − ζ ( 4 ) = 6 1 π 2 G − 9 0 1 π 4 where G is Catalan's constant. Putting this all together gives b = 1 ∑ ∞ b 2 coth 2 π b = 3 2 G + 1 8 0 1 9 π 2 making the answer 2 + 3 + 1 9 + 1 8 0 + 2 = 2 0 6 .