练习 14.3
T1
设 \(\alpha>0\) ，求证：
$$ \sum\limits_{n=1}^{\infty} \left|\dfrac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{n!}\right| $$
收敛.
考虑 Raabe 判别法：
\[
\frac{u_{n}}{u_{n+1}} = n\frac{\alpha+1}{n-\alpha}
\]
其中 \(n\) 取到大于 \([\alpha]\) 的值，此时有
\[
\lim_{n\to \infty} n \left(\frac{u_{n}}{u_{n+1}}-1\right) = \lim_{n\to \infty} n\frac{\alpha+1}{n-\alpha} = \alpha+1 >1
\]
根据 Raabe 判别法可知该级数收敛. \(\square\)
T2
讨论下列级数的收敛性：
(1) \(\displaystyle\sum\limits_{n=1}^{\infty} \dfrac{\sqrt{n!}}{(a+1)(a+\sqrt{2})\cdots (a+\sqrt{n})},a>0\).
(2) \(\displaystyle\sum\limits_{n=1}^{\infty} \dfrac{n!n^{-p}}{q(q+1)\cdots(q+n)},p>0,q>0\).
(1) 考虑
\[
n\left(\dfrac{u_{n}}{u_{n+1}}-1\right) = \frac{an}{\sqrt{n+1}}\to +\infty
\]
由 Raabe 判别法可知该级数收敛.
(2) 先利用 Raabe 判别法，计算有
\[
\begin{aligned}
\frac{u_{n}}{u_{n+1}} & = (q+n+1) \dfrac{(n+1)^{p}}{(n+1)n^{p}} \\
& = \left(1+\frac{1}{n}\right)^{p} + \frac{q(n+1)^{p-1}}{n^{p}} \\
\end{aligned}\tag{2.1}
\]
那么相应的
\[
\begin{aligned}
\lim_{n\to \infty}n\left(\frac{u_{n}}{u_{n+1}}-1\right) & = \lim_{n\to \infty}n\left[\left(1+\frac{1}{n}\right)^{p} + \frac{q(n+1)^{p-1}}{n^{p}}-1\right] \\
& = \lim_{n\to \infty} n\left[\left(1+ \frac{1}{n}\right)^{p}-1\right] + \lim_{n\to \infty} q\cdot\dfrac{(n+1)^{p-1}}{n^{p-1}} \\
& = p+q
\end{aligned}
\]
因此根据 Raabe 判别法，当 \(p+q<1\) 时级数发散，\(p+q>1\) 时级数收敛，那么最后仅需讨论 \(p+q=1\) 时的敛散性.
在 \(p+q=1\) 时，对 (2.1) 式利用 Newton 二项式展开：
\[
\begin{aligned}
(2.1) & = \left(1+\dfrac{1-p}{n+1}\right) \left[1+ \frac{p}{n}+ \frac{p(p-1)}{2n^{2}}+ o\left(\frac{1}{n^{2}}\right)\right] \\
& = 1+ \frac{1-p}{n+1} + \frac{p}{n} + \dfrac{p(1+p)}{n(n+1)} + \frac{p(p-1)}{2n^{2}} + o\left(\frac{1}{n^{2}}\right)
\end{aligned}
\]
利用 Gauss 判别法即有该级数发散. 因此，该级数收敛当且仅当 \(p+q >1\) . \(\square\)
Remark.
具体的正项级数收敛性问题：比值相关判别法的进化过程是：D'Alembert -> Raabe -> Gauss -> Bertrand.
比值所能知道的：
\[
\frac{u_{n}}{u_{n+1}} = a_{0} + \frac{a_{1}}{n} + \frac{a_{2}}{n(\ln n)} + \frac{a_{3}}{n\ln n\ln\ln n} + \cdots
\]