Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/242




 * Following a line of reasoning similar to that applied to (A) and (B), it is evident that, if


 * {| style="width:100%"


 * align="center" | $$ u_1 + u_2 + u_3 + \cdots$$
 * }
 * }


 * is a series of positive terms to be tested, which are never less than the corresponding terms of the series of positive terms, namely,


 * {| style="width:100%"


 * align="center" | $$ b_1 + b_2 + b_3 + \cdots$$
 * }
 * }


 * known to be divergent, then (E) is a divergent series.

. Test the series

$$1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{4}} + \cdots $$.

Solution. This series is divergent, since its terms are greater than the corresponding terms of the harmonic series $$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots $$. which is known (§137) to be divergent.

. Test the following series (called the p series) for different values of p:
 * {| style="width:100%"


 * align="center" | $$ 1 + \frac{1}{2^p} + \frac{1}{3^p} + \frac{1}{4^p} + \cdots $$
 * }
 * }

Solution. Grouping the terms, we have, when $$p > 1$$,

$$\begin{align} \frac{1}{2^p} + \frac{1}{3^p} < \frac{1}{2^p} + \frac{1}{2^p} = \frac{2}{2^p} & = \frac{1}{2^{p-1}} \\ \frac{1}{4^p} + \frac{1}{5^p} + \frac{1}{6^p} + \frac{1}{7^p} < \frac{1}{4^p} + \frac{1}{4^p} + \frac{1}{4^p} + \frac{1}{4^p} = \frac{4}{4^p} & = \left ( \frac{1}{2^{p-1}} \right )^2 \\ \frac{1}{8^p} + \cdots + \frac{1}{15^p} < \frac{1}{8^p} + \frac{1}{8^p} + \frac{1}{8^p} + \frac{1}{8^p} + \frac{1}{8^p} + \frac{1}{8^p} + \frac{1}{8^p} + \frac{1}{8^p} = \frac{8}{8^p} & = \left ( \frac{1}{2^{p-1}} \right )^3 \\ \end{align}$$

and so on. Construct the series

When $$p > 1 $$, series (H) is a geometric series with the common ration less than unity, and is therefore convergent. But the sum of (G) is less than the sum of (H), as shown in the above inequalities; therefore  (G) is also convergent.

When $$p = 1$$ series (G) becomes the harmonic series which we saw was divergent, and neither of the above tests apply.

When $$p < 1$$, the terms of series (G) will, after the first, be greater than the corresponding terms of the harmonic series; hence (G) is divergent.

Let

be a series of positive termes to be tested.