Editing Sinc function (section)

== Summation ==
All sums in this section refer to the unnormalized sinc function.

The sum of {{math|sinc(''n'')}} over integer {{mvar|n}} from 1 to {{math|∞}} equals {{math|{{sfrac|{{pi}} − 1|2}}}}:

<math display="block">\sum_{n=1}^\infty \operatorname{sinc}(n) = \operatorname{sinc}(1) + \operatorname{sinc}(2) +  \operatorname{sinc}(3) + \operatorname{sinc}(4) +\cdots = \frac{\pi - 1}{2}.</math>

The sum of the squares also equals {{math|{{sfrac|{{pi}} − 1|2}}}}:<ref>{{cite journal | title = Advanced Problem 6241 | journal = American Mathematical Monthly | date = June–July 1980 | volume = 87 | issue = 6 | pages = 496–498 | publisher = [[Mathematical Association of America]] | location = Washington, DC | doi = 10.1080/00029890.1980.11995075}}</ref><ref name="BBB">{{cite journal | author1=Robert Baillie | author2-link=David Borwein | author2=David Borwein | author3=Jonathan M. Borwein | author3-link=Jonathan M. Borwein | title=Surprising Sinc Sums and Integrals | journal=American Mathematical Monthly | date=December 2008 | volume=115 | issue=10 | pages=888–901 | jstor = 27642636 | doi=10.1080/00029890.2008.11920606 | hdl=1959.13/940062 | s2cid=496934 | hdl-access=free}}</ref>

<math display="block">\sum_{n=1}^\infty \operatorname{sinc}^2(n) = \operatorname{sinc}^2(1) + \operatorname{sinc}^2(2) + \operatorname{sinc}^2(3) + \operatorname{sinc}^2(4) + \cdots = \frac{\pi - 1}{2}.</math>

When the signs of the [[addend]]s alternate and begin with +, the sum equals {{sfrac|1|2}}:
<math display="block">\sum_{n=1}^\infty (-1)^{n+1}\,\operatorname{sinc}(n) = \operatorname{sinc}(1) - \operatorname{sinc}(2) + \operatorname{sinc}(3) - \operatorname{sinc}(4) + \cdots = \frac{1}{2}.</math>

The alternating sums of the squares and cubes also equal {{sfrac|1|2}}:<ref name="FWFS">{{cite arXiv |last=Baillie |first=Robert |eprint=0806.0150v2 |class=math.CA |title=Fun with Fourier series |date=2008}}</ref>
<math display="block">\sum_{n=1}^\infty (-1)^{n+1}\,\operatorname{sinc}^2(n) = \operatorname{sinc}^2(1) - \operatorname{sinc}^2(2) + \operatorname{sinc}^2(3) - \operatorname{sinc}^2(4) + \cdots = \frac{1}{2},</math>

<math display="block">\sum_{n=1}^\infty (-1)^{n+1}\,\operatorname{sinc}^3(n) = \operatorname{sinc}^3(1) - \operatorname{sinc}^3(2) + \operatorname{sinc}^3(3) - \operatorname{sinc}^3(4) + \cdots = \frac{1}{2}.</math>