Give a concise proof that, asymptotically as n grows,

analysis/alternating.series New

New entry, edited from material collected for the archive and not yet in a released version.

Problem

Give a concise proof that, asymptotically as n grows,

  1/(n+1) - 1/(n+3) + 1/(n+5) - 1/(n+7) + ...  ~

              1/(2n) - 1/(2n^3) + 5/(2n^5) - 61/(2n^7) + 1385/(2n^9) - ...

where 1, -1, 5, -61, 1385, ... are the Euler numbers.  (The series on
the right does not converge, but the error is always less than the
first term omitted.)
[ Editor: problem statement reconstructed from C. E. Thompson's posting
quoted in the solution. ]

Show solution

In article <1992Nov4.142040.5774@infodev.cam.ac.uk>, cet1@cus.cam.ac.uk
(C.E. Thompson) writes:
>However, I wouldn't like to show you my current proof that the
>Euler-Maclaurin formula reduces to this simple form for this series,
>as it is quite horrible.
>Can anyone produce a concise and/or elegant proof?

I'll use operators to make things concise and elegant; D is the
derivative operator, and E is the shift operator.  In this notation,
Taylor's theorem is:

E = exp D

You can do a bit of algebra and get the following:

(1 - E)^-1 = 1/2 (1 - coth (1/2 D))

This is the Euler-Maclaurin formula.  But it's not really what we
want.  The series should be alternating, and the sample points should
be offset from the derivatives.  We can try the following one:

E (1 + E^2)^-1 = 1/2 sech D

I'll expand this out in case you don't follow the operator stuff:

  inf                           inf
 ----                          ----   E
 \        j                  1 \       j    (2 j)
  >   (-1)  f(n + 2 j + 1) = -  >   ------ f     (n)
 /                           2 /    (2 j)!
 ----                          ----
 j = 0                         j = 0

The Euler numbers come from the secant series, just like the Bernoulli
numbers come from the tangent series in the Euler-Maclaurin formula.  I
don't know if this formula has a specific name; I just consider it a
variant of the Euler-Maclaurin formula.

You can play around with this formula, trying out different functions
for f.  It usually works, even though both series often don't
converge.  If f is a polynomial, the series on the right terminates, so
this is easy to test.  You can even derive it this way, since the Euler
numbers can be defined in terms of differences of polynomials.

In the case we're talking about, f(n) is n^-1, so f^(j)(n) is
j! n^-(j+1).  The factorials cancel and we're left with just the Euler
numbers.  This gives exactly the formula above for the error in the
Leibniz series.

-- Joe Keane <jgk@osc.com>

[ Editor: strictly f^(j)(n) = (-1)^j j! n^-(j+1); only even j occur
in the formula, so the sign does not matter here. ]