Function: sumnumlagrangeinit
Section: sums
C-Name: sumnumlagrangeinit
Prototype: DGDGp
Help: sumnumlagrangeinit({asymp},{c1}): initialize tables for Lagrange
 summation of a series.
Doc: initialize tables for Lagrange summation of a series. By
 default, assume that the remainder $R(n) = \sum_{m \geq n} f(m)$
 has an asymptotic expansion
 $$R(n) = \sum_{m \geq n} f(n) \approx \sum_{i\geq 1} a_{i} / n^{i}$$
 at infinity. The argument \kbd{asymp} allows to specify different
 expansions:

 \item a real number $\beta$ means
 $$ R(n) = n^{-\beta} \sum_{i\geq 1} a_{i} / n^{i} $$

 \item a \typ{CLOSURE} $g$ means
 $$R(n) = g(n) \sum_{i\geq 1} a_{i} / n^{i}$$
 (The preceding case corresponds to $g(n) = n^{-\beta}$.)

 \item a pair $[\alpha,\beta]$ where $\beta$ is as above and
 $\alpha\in \{2, 1, 1/2, 1/3, 1/4\}$. We let $R_{2}(n) = R(n) - f(n)/2$
 and $R_{\alpha}(n) = R(n)$ for $\alpha\neq 2$. Then
 $$R_{\alpha}(n) = g(n) \sum_{i\geq 1} a_{i} / n^{i\alpha}$$
 Note that the initialization times increase considerable for the $\alpha$
 is this list ($1/4$ being the slowest).

 The constant $c1$ is technical and computed by the program, but can be set
 by the user: the number of interpolation steps will be chosen close to
 $c1\cdot B$, where $B$ is the bit accuracy.

 \bprog
 ? \p2000
 ? sumnumlagrange(n=1, n^-2);
 time = 173 ms.
 ? tab = sumnumlagrangeinit();
 time = 172 ms.
 ? sumnumlagrange(n=1, n^-2, tab);
 time = 4 ms.

 ? \p115
 ? sumnumlagrange(n=1, n^(-4/3)) - zeta(4/3);
 %1 = -0.1093[...] \\ junk: expansion in n^(1/3)
 time = 84 ms.
 ? tab = sumnumlagrangeinit([1/3,0]); \\ alpha = 1/3
 time = 336 ms.
 ? sumnumlagrange(n=1, n^(-4/3), tab) - zeta(4/3)
 time = 84 ms.
 %3 = 1.0151767349262596893 E-115 \\ now OK

 ? tab = sumnumlagrangeinit(1/3); \\ alpha = 1, beta = 1/3: much faster
 time = 3ms
 ? sumnumlagrange(n=1, n^(-4/3), tab) - zeta(4/3) \\ ... but wrong
 %5 = -0.273825[...]   \\ junk !
 ? tab = sumnumlagrangeinit(-2/3); \\ alpha = 1, beta = -2/3
 time = 3ms
 ? sumnumlagrange(n=1, n^(-4/3), tab) - zeta(4/3)
 %6 = 2.030353469852519379 E-115 \\ now OK
 @eprog\noindent in The final example with $\zeta(4/3)$, the remainder
 $R_{1}(n)$ is of the form $n^{-1/3} \sum_{i\geq 0} a_{i} / n^{i}$, i.e.
 $n^{2/3} \sum_{i\geq 1} a_{i} / n^{i}$. The explains the wrong result
 for $\beta = 1/3$ and the correction with $\beta = -2/3$.
