Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/246

	or compare the given series with some series which is known to be divergent, as
	$1+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}+\cdots ;$	(harmonic series)
	$1+{\frac {1}{2^{p}}}+{\frac {1}{3^{p}}}+{\frac {1}{4^{p}}}+\cdots ;p<1$	(p series)

Illustrative Example 1. Test the series

$1+{\frac {1}{1!}}+{\frac {1}{2!}}+{\frac {1}{3!}}+{\frac {1}{4!}}+\cdots$

Solution. Here

$u_{n}={\frac {1}{\left(n-1\right)!}},\;u_{n+1}={\frac {1}{n!}}$

$\therefore \lim _{n\to \infty }\left({\frac {u_{n+1}}{u_{n}}}\right)=lim_{n\to \infty }\left({\frac {\frac {1}{n}}{\frac {1}{\left(n-1\right)!}}}\right)=\lim _{n\to \infty }\left({\frac {\left(n-1\right)!}{n!}}\right)=\lim _{n\to \infty }\left({\frac {1}{n}}\right)=0\left(=p\right)$

and by I, §141, the series is convergent.

Illustrative Example 2. Test the series

${\frac {1!}{10}}+{\frac {2!}{10^{2}}}+{\frac {3!}{10^{3}}}+\cdots$

Solution. Here

$u_{n}={\frac {n!}{10^{n}}},\;u_{n+1}={\frac {\left(n+1\right)!}{10^{n+1}}}$

$\therefore \lim _{n\to \infty }\left({\frac {u_{n+1}}{u_{n}}}\right)=\lim _{n\to \infty }\left({\frac {\left(n+1\right)!}{10^{n+1}}}\times {\frac {10^{n}}{n!}}\right)=\lim _{n\to \infty }\left({\frac {n+1}{10}}\right)=\infty \left(=p\right)$

and by II,§141 the series is divergent.

Illustrative Example 3. Test the series

(C)	${\frac {1}{1\cdot 2}}+{\frac {1}{3\cdot 4}}+{\frac {1}{5\cdot 6}}+\cdots$

Solution. Here

$u_{n}={\frac {1}{\left(2n-1\right)2n}},\;u_{n+1}{\frac {1}{\left(2n+1\right)\left(2n+2\right)}}$

$\therefore \lim _{n\to \infty }\left({\frac {u_{n+1}}{u_{n}}}\right)=\lim _{n\to \infty }\left[{\frac {\left(2n-1\right)2n}{\left(2n+1\right)\left(2n+2\right)}}\right]={\frac {\infty }{\infty }}$

This being an indeterminate form, we evaluate it, using the rule in §112.

Differentiating,

$\lim _{n\to \infty }\left({\frac {8n-2}{8n-6}}\right)={\frac {\infty }{\infty }}$

Differentiating again,

$\lim _{n\to \infty }\left({\frac {8}{8}}\right)=1\left(=p\right)$

This gives no test (III, §141). But if we compare series (C) with (G), §138,making $p=2$, namely,

(D)	${\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}+{\frac {1}{4^{2}}}+\cdots$

we see that (C) must be convergent, since its terms are less than the corresponding terms of (D), which was proved convergent.