Editing Euler–Maclaurin formula (section)

===Asymptotic expansion of sums===
In the context of computing [[asymptotic expansion]]s of sums and [[Series (mathematics)|series]], usually the most useful form of the Euler–Maclaurin formula is
<math display=block>\sum_{n=a}^b f(n) \sim \int_a^b f(x)\,dx + \frac{f(b) + f(a)}{2} + \sum_{k=1}^\infty \,\frac{B_{2k}}{(2k)!} \left(f^{(2k - 1)}(b) - f^{(2k - 1)}(a)\right),</math>

where {{mvar|a}} and {{mvar|b}} are integers.<ref>{{Cite book | editor1-last=Abramowitz | editor1-first=Milton | editor1-link=Milton Abramowitz | editor2-last=Stegun | editor2-first=Irene A. | editor2-link=Irene Stegun | title=[[Abramowitz and Stegun|Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables]] | publisher=[[Dover Publications]] | location=New York | isbn=978-0-486-61272-0 | year=1972 | pages = 16, 806, 886}}</ref> Often the expansion remains valid even after taking the limits {{math|''a'' → −∞}} or {{math|''b'' → +∞}} or both.  In many cases the integral on the right-hand side can be evaluated in [[Differential Galois theory|closed form]] in terms of [[elementary function]]s even though the sum on the left-hand side cannot. Then all the terms in the asymptotic series can be expressed in terms of elementary functions.  For example,
<math display=block>\sum_{k=0}^\infty \frac{1}{(z + k)^2} \sim \underbrace{\int_0^\infty\frac{1}{(z + k)^2}\,dk}_{= \dfrac{1}{z}} + \frac{1}{2z^2} + \sum_{t = 1}^\infty \frac{B_{2t}}{z^{2t + 1}}.</math>

Here the left-hand side is equal to {{math|''ψ''<sup>(1)</sup>(''z'')}}, namely the first-order [[Polygamma function#Series representation|polygamma function]] defined by
:<math>\psi^{(1)}(z) = \frac{d^2}{dz^2}\log \Gamma(z);</math>
the [[gamma function]] {{math|Γ(''z'')}} is equal to {{math|(''z'' − 1)!}} when {{mvar|z}} is a [[positive integer]]. This results in an asymptotic expansion for {{math|''ψ''<sup>(1)</sup>(''z'')}}.  That expansion, in turn, serves as the starting point for one of the derivations of precise error estimates for [[Stirling's approximation]] of the [[factorial]] function.

====Examples====
If {{mvar|s}} is an integer greater than 1 we have:
<math display=block>\sum_{k=1}^n \frac{1}{k^s} \approx \frac 1{s-1}+\frac 12-\frac 1{(s-1)n^{s-1}}+\frac 1{2n^s}+\sum_{i=1}\frac{B_{2i}}{(2i)!}\left[\frac{(s+2i-2)!}{(s-1)!}-\frac{(s+2i-2)!}{(s-1)!n^{s+2i-1}}\right].</math>

Collecting the constants into a value of the [[Riemann zeta function]], we can write an asymptotic expansion:
<math display=block>\sum_{k=1}^n \frac{1}{k^s} \sim\zeta(s)-\frac 1{(s-1)n^{s-1}}+\frac 1{2n^s}-\sum_{i=1}\frac{B_{2i}}{(2i)!}\frac{(s+2i-2)!}{(s-1)!n^{s+2i-1}}.</math>

For {{mvar|s}} equal to 2 this simplifies to
<math display=block>\sum_{k=1}^n \frac{1}{k^2} \sim\zeta(2)-\frac 1n+\frac 1{2n^2}-\sum_{i=1}\frac{B_{2i}}{n^{2i+1}},</math>
or
<math display=block>\sum_{k=1}^n \frac{1}{k^2} \sim \frac{\pi^2}{6} -\frac{1}{n} +\frac{1}{2n^2} -\frac{1}{6n^3}+\frac{1}{30n^5}-\frac{1}{42n^7} + \cdots.</math>

When {{math|''s'' {{=}} 1}}, the corresponding technique gives an asymptotic expansion for the [[harmonic number]]s:
<math display=block>\sum_{k=1}^n \frac{1}{k} \sim \gamma + \log n + \frac{1}{2n} - \sum_{k=1}^\infty \frac{B_{2k}}{2kn^{2k}},</math>
where {{math|''γ'' ≈ 0.5772...}} is the [[Euler–Mascheroni constant]].