Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/243

Divide any general term by the one that immediately precedes it; i.e. form the test ratio ${\tfrac {u_{n+1}}{u_{n}}}$.

As $n$ increases without limit, let $\lim _{n\to \infty }{\tfrac {u_{n+1}}{u_{n}}}=p$.

I. When $p<1$. By the definition of a limit (§13) we can choose n so large, say $n=m$, that when $n\geqq m$ the ratio ${\tfrac {u_{n+1}}{u_{n}}}$ shall differ from $p$ by as little as we please, and therefore be less than a proper fraction $r$. Hence

$u_{m+1}<u_{m}r;\;u_{m+2}<u_{m+1}r<u_{m}r^{2};\;u_{m+3}r<u_{m}r^{3}$

and so on. Therefore, after the term $u_{m}$ , each term of the series (A) is less than the corresponding term of the geometrical series

(B)	$u_{m}r+u_{m}r^{2}+u_{m}r^{3}+\cdots$

But since $r<1$, the series (B), and therefore also the series is convergent.^[1]

II. When $p>1$ (or $p=\infty$). Following the same line of reasoning as in I, the series (A) may be shown to be divergent.

III. When $p=1$, the series maybe either convergent or divergent; that is, there is no test. For, consider the $p$ series, namely,

$1+{\frac {1}{2^{p}}}+{\frac {1}{3^{p}}}+{\frac {1}{4^{p}}}+\cdots +{\frac {1}{n^{p}}}+{\frac {1}{\left(n+1\right)^{p}}}+\cdots$

The test ratio is

${\frac {u_{n+1}}{u_{n}}}=\left({\frac {n}{n+1}}\right)^{p}=\left(1-{\frac {1}{n+1}}\right)^{p}$

and

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

Hence $p=1$, no matter what value $p$ may have. But in §138 we showed that

when $p>1$ the series converges, and

when $p\leqq 1$, the series diverges.

Thus it appears that $p$ can equal unity both for convergent and for divergent series, and the ratio test for convergence fails. There are other tests to apply in cases like this, but the scope of our book does not admit of their consideration.

1. When examining a series for convergence we are at liberty to disregard any finite number of terms; the rejection of terms would affect the value but not the existence of the limit.