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 $$\lim_{n \to \infty} \tfrac{u_{n+1}}{u_n} =p$$.

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

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

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

When $$p > 1$$ (or $$p = \infty$$). Following the same line of reasoning as in I, the series (A) may be shown to be divergent. . When $$p = 1$$, the series maybe either convergent or divergent; that is, there is no test. For, consider the $$p$$ series, namely,

The test ratio is

and

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.