§ 3. Bounds for the remainder in the euler-maclaurin summation formula

1. Bounds for the remainder in the euler-maclaurin summation formula

The estimate obtained in (16) for the Bernoulli polynomials on the interval $[0, 1]$ allows one to estimate the remainder $T_p(x, n)$ in the Euler-Maclaurin summation formula (VI, p. 282, formula (39)) easily:

\left\{ \begin{array}{l} f(x) + f(x+1) + \cdots + f(x+n) \\ \phantom{=} = \int_x^{x+n+1} f(t)\,dt - \frac{1}{2} (f(x+n+1) - f(x)) \\ \phantom{=} \phantom{=} + \sum_{k=1}^p \frac{b_{2k}}{(2k)!} (f^{(2k-1)}(x+n+1) - f^{(2k-1)}(x)) + T_p(x, n). \end{array} \right.

Indeed, one has (VI, p. 283, formula (41))

T_p(x, n) = -\frac{1}{(2p+1)!} \int_0^{n+1} \overline{B}_{2p+1}(1-s) f^{(2p+1)}(x+s)\,ds

where $\overline{B}_{2p+1}(t)$ is the periodic function of period 1 equal to $B_{2p+1}(t)$ on the interval $[0, 1[$ . The formula (16) of VI, p. 288, shows that

|\overline{B}_{2p+1}(t)| \leq 4e^{2\pi} \frac{(2p+1)!}{(2\pi)^{2p+1}}

(3)

for all $t \in \mathbf{R}$ , and applying the mean value formula gives the estimate

|T_p(x, n)| \leq \frac{4e^{2\pi}}{(2\pi)^{2p+1}} \int_x^{x+n+1} |f^{(2p+1)}(t)| \, dt

(4)

for $T_p(x, n)$ .

2. Application to asymptotic expansions

The Euler-Maclaurin formula allows one to give a more complete solution (in the most important cases) to the problem treated in V, p. 238 to 242, that of obtaining an asymptotic expansion of the partial sum $s_n = \sum_{m=0}^n g(m)$ (resp. of the remainder $r_n = \sum_{m=n+1}^\infty g(m)$ ), where $g$ is a scalar function, $> 0$ , monotone, defined on $[0, +\infty[$ .

We shall restrict ourselves to the case where $g$ is an (H) function (V, p. 252), of order 0 relative to $e^x$ ; in other words, one has the relation $g' \ll g$ ; from this relation one deduces $g^{(k+1)} \ll g^{(k)}$ for every integer $k > 0$ such that none of the derivatives $g^{(h)}$ of order $h \leq k$ is equivalent to a constant (V, p. 232, prop. 7). Let $p$ be an integer such that none of the derivatives $g^{(h)}$ of order $h \leq 2p$ is equivalent to a constant. First we assume that the series with general term $g(n)$ has infinite sum, and distinguish several cases:

1. $|g^{(2p-1)}(n)|$ tends to $+\infty$ with $n$ ; we have the same, by the hypothesis, for $|g^{(2k-1)}(n)|$ for $1 \leq k \leq p$ ; further, since $g^{(2p+1)}$ is monotone on a neighbourhood of $+\infty$ , the formula (4) of VI, p. 289, gives $T_p(0, n) = O(g^{(2p)}(n+1)) = o(g^{(2p-1)}(n+1))$ ; the Euler-Maclaurin formula, applied for $x = 0$ , shows that

s_n = \sum_{m=0}^n g(m) = \int_0^{n+1} g(t) \, dt - \frac{1}{2} g(n+1)

+ \sum_{k=1}^p \frac{b_{2k}}{(2k)!} g^{(2k-1)}(n+1) + o(g^{(2p-1)}(n+1))

each of the terms of this sum being negligible relative to the preceding one; on expanding each of them relative to a comparison scale $\mathscr{E}$ one will then have an asymptotic expansion for $s_n$ .

2) Now suppose that for an index $q$ such that $1 \leq q \leq p$ we have $|g^{(2q-1)}(n)|$ tending to $+\infty$ with $n$ , but that $g^{(2k-1)}(n)$ tends to 0 for $k > q$ . Since $g^{(2p+1)}$ is monotone on a neighbourhood of $+\infty$ the integral $\int_0^\infty |g^{(2p+1)}(u)| \, du$ converges, and one can then write

s_n = \sum_{m=0}^{n} g(m) = \int_0^{n+1} g(t)\,dt - \frac{1}{2}g(n+1) + \sum_{k=1}^{q} \frac{b_{2k}}{(2k)!}\,g^{(2k-1)}(n+1) + C

+ \sum_{k=q+1}^{p} \frac{b_{2k}}{(2k)!}\,g^{(2k-1)}(n+1) + o\left(g^{(2p-1)}(n+1)\right)

where C is a constant: indeed

\int_{n+1}^{\infty} |g^{(2p+1)}(u)|\,du = O\left(g^{(2p)}(n+1)\right) = o\left(g^{(2p-1)}(n+1)\right).

The same formula is valid when $g(n)$ itself tends to 0. Finally, when the series with general term $g(n)$ converges one has, for the remainder $r_n = \sum_{m=n+1}^{\infty} g(m)$ , the expansion

r_n = \sum_{m=n+1}^{\infty} g(m) = \int_{n+1}^{\infty} g(t)\,dt + \frac{1}{2}\,g(n+1)

- \sum_{k=1}^{p} \frac{b_{2k}}{(2k)!}\,g^{(2k-1)}(n+1) + o\left(g^{(2p-1)}(n+1)\right).

Exercises

See the exercises for § 3.