May 14, 2012
Cute sums
Here are two cute sums:
(e+1)/2(e-1) = Σn=-∞∞ 1/(1 + 4 π2 n2)
π/(1-exp(-π)) - π/2 = Σn=-∞∞ 1/(1+4 n2)
The way to prove it is to use the Poisson summation formula, which states that for appropriate functions:
Σn=-∞∞ f(n) = Σn=-∞∞ F(n)
where F is the Fourier transform of f. The "appropriate" part is of course where tricky things may happen, since one or both sums might not converge.
Now, f(x) = exp(-a|x|) has the transform F(ξ) = 2a/(a2 + 4 π2 ξ2) for a>0.
So we get this possible equality:
Σn=-∞∞ exp(-a|x|) = Σn=-∞∞ 2a/(a2 + 4 π2 n2)
These sums are both well behaved, since both can easily be bounded by nicely convergent integrals, so convergence is assured.
The left hand side can be rewritten as [2 Σ_n=0∞ exp(-ax)] - 1 since exp(-a|x|) is an even function (imagine "folding" the number line around n=0; the -1 term corrects the double counting of the n=0 term with value 1). And the sum inside the brackets is a nice geometric series summing to 1/(1-exp(-a)). So the original sum is
2/(1-exp(-a)) - 1 = Σn=-∞∞ 2a/(a2 + 4 π2 \n2)
Some rearrangement produces
1/a(1-exp(-a)) - 1/2a = Σn=-∞∞ 1/(a2+4π2 n2)
Now, we can set a=1 and get
1/(1-e^-1) - 1/2 = Σ_n=-infty∞ 1/(1 + 4 π2 n2)
a=π gives
π/(1-exp(-π)) - π/2 = Σn=-∞∞ 1/(1+4 n2)
Maybe not terribly useful, but beautiful. I remember how delightful it was to learn how to calculate integrals using residues or exploit Fourier series to get cool sums and integrals in closed form. After all, the coolest fact about
π/3 = Σn=1∞ sin(nπ/3)/n
is that it is so easy to derive - once you have the powerful machinery behind it.
Addendum 15/5/12:
Just found another nice identity the same way, this time using the transform pair f(x)=x H(x) exp(-αx), F(ξ)=1/(α+2πξi)2 where H(x) is the Heaviside step function. In the same way as above this leads to:
4π2Σn=0∞ n exp(-2πn) = Σn=-∞∞ (1-n2)/(1+n2)2
A neat thing about it is that the sum on the left is all positive, while the sum on the right has just one positive term.
Posted by Anders3 at May 14, 2012 05:28 PM