Proposition 6

Let g be a real function, > 0 and monotone, defined on an interval [x0, +\infty[ (where x_0 \leq 1 ), and such that log g and x are comparable to order 1.*

1° If g is of infinite order relative to e, one has

2° If g is of finite order $\mu$ with respect to $e^x$ one has

(the number $\frac{\mu}{1 - e^{-\mu}}$ is to be replaced by 1 in (3) and (4) when $\mu = 0$).

1° If g is of order $+\infty$ relative to $e^x$ one has $\log g \gg x$ whence $g'/g \gg 1$, or $g' \gg g$, by the hypothesis; g is therefore increasing and tends to $+\infty$ with $x$, whence $\sum_{n=1}^\infty g(n) = +\infty$. If $u$ is the step function associated with the series $(g(n))$ (V, p. 237), one has $u(x) \leq g(x)$ starting from a certain value of $x$, so $u \ll g$ and consequently

since $s_n = s_{n-1} + g(n)$ one has $s_n \sim g(n)$. The proof is similar when g is of order $-\infty$ relative to $e^x$; we thus obtain formula (2).

2° If g is of finite order $\mu$ relative to $e^x$ one can write $g(x) = e^{\mu x} h(x)$ where h is of order 0 relative to $e^x$; further, by hypothesis, $\log g \sim \mu x$ for $\mu \neq 0$ ($\log g \ll x$ for $\mu = 0$) implies $h' \ll h$. Suppose first that $\sum_{n=1}^\infty g(n) = +\infty$ (which implies that $\mu \geq 0$; the converse always holds if $\mu > 0$, since then $g(x)$ tends to $+\infty$ with $x$); let us evaluate the principal part of $\int_{n-1}^n g(t) \, dt$. One can write

Now, the relation $h' \ll h$ implies that for every $\varepsilon > 0$ there is an $n_0$ such that the relation $x \geq n_0$ implies that $|h'(x)/h(x)| \leq \varepsilon$; one deduces from the mean value theorem that $-\varepsilon \leq \log |h(t)/h(n)| \leq \varepsilon$, for $n-1 \leq t \leq n$, if $n \geq n_0$, whence

and consequently

since $e^{\mu t}$ is increasing. Since $e^\varepsilon - 1$ becomes arbitrarily small with $\varepsilon$ one sees that one can write

$\left( \frac{1 - e^{-\mu}}{\mu} \text{ being replaced by } 1 \text{ when } \mu = 0 \right)$. The proposition is then a consequence of prop. 2 of V, p. 237. One argues similarly when $\sum_{n=1}^\infty g(n)$ is finite.

By applying prop. 6 of V, p. 239, repeatedly one can then sometimes obtain an asymptotic expansion for $s_n = \sum_{m=1}^n g(m)$. First suppose that $g$ is of order $+\infty$ relative to $e^\lambda$; for every fixed value of $p$ one can write, by prop. 6,

and it suffices to expand (relative to $\mathscr{E}'$) each of the functions $g(n-k)$ ($0 \leq k \leq p$), limiting the precision of the expansions to the principal part of $g(n-p)$, to obtain an expansion for the $s_n$.