Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/241



We notice that every term of (G) is equal to or less than the corresponding term of (F), so that the sum of any number of the first terms of (F) will be greater than the sum of the corresponding terms of (G). But since the sum of the terms grouped in each bracket in (G) equals $$\tfrac{1}{2}$$, the sum of (G) may be made as large as we please by taking terms enough. The sum (G) increases indefinitely as the number of terms increases without limit; hence (G), and therefore also (F), is divergent.

We shall now proceed to deduce special tests which, as a rule, are easier to apply than the above theorems.

In many cases, an example of which was given in the last section, it is easy to determine whether or not a given series is convergent by comparing it term by term with another series whose character is known. Let

''be a series of positive terms which it is desired to test for convergence. If a series of positive terms already known to be convergent, namely,''

can be found whose terms are never less than the corresponding terms in the series (A) to be tested, then (A) is a convergent series and its sum does not exceed that of (B).

Proof. Let and and suppose that Then, since it follows that $$s_n < A$$. Hence, by Theorem I, §136, $$s_n$$ approaches a limit; therefore the series (A) is convergent and the limit of its sum is not greater than $$A$$.

Test the series
 * {| style="width:100%"


 * align="center" | $$1 + \frac{1}{2^2} + \frac{1}{3^3} + \frac{1}{4^4} + \frac{1}{5^5} + \cdots$$.
 * }
 * }


 * Solution. Each term after the first is less than the corresponding term of the geometric series


 * {| style="width:100%"


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


 * which is known to be convergent; hence (C) is also convergent.