Following a line of reasoning similar to that applied to (A ) and (B ) , it is evident that, if
(E )
u
1
+
u
2
+
u
3
+
⋯
{\displaystyle 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,
(F )
b
1
+
b
2
+
b
3
+
⋯
{\displaystyle b_{1}+b_{2}+b_{3}+\cdots }
known to be divergent, then (E) is a divergent series. Illustrative Example 2. . Test the series
1
+
1
2
+
1
3
+
1
4
+
⋯
{\displaystyle 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
+
1
2
+
1
3
+
1
4
+
⋯
{\displaystyle 1+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}+\cdots }
.
which is known (§137 ) to be divergent.
Illustrative Example 3. . Test the following series (called the p series) for different values of p:
(G )
1
+
1
2
p
+
1
3
p
+
1
4
p
+
⋯
{\displaystyle 1+{\frac {1}{2^{p}}}+{\frac {1}{3^{p}}}+{\frac {1}{4^{p}}}+\cdots }
Solution. Grouping the terms, we have, when
p
>
1
{\displaystyle p>1}
1
2
p
+
1
3
p
<
1
2
p
+
1
2
p
=
2
2
p
=
1
2
p
−
1
1
4
p
+
1
5
p
+
1
6
p
+
1
7
p
<
1
4
p
+
1
4
p
+
1
4
p
+
1
4
p
=
4
4
p
=
(
1
2
p
−
1
)
2
1
8
p
+
⋯
+
1
15
p
<
1
8
p
+
1
8
p
+
1
8
p
+
1
8
p
+
1
8
p
+
1
8
p
+
1
8
p
+
1
8
p
=
8
8
p
=
(
1
2
p
−
1
)
3
{\displaystyle {\begin{aligned}{\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{aligned}}}
and so on. Construct the series
(H )
1
+
1
2
p
−
1
+
(
1
2
p
−
1
)
2
+
(
1
2
p
−
1
)
3
+
⋯
{\displaystyle 1+{\frac {1}{2^{p-1}}}+\left({\frac {1}{2^{p-1}}}\right)^{2}+\left({\frac {1}{2^{p-1}}}\right)^{3}+\cdots }
When
p
>
1
{\displaystyle p>1}
(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
{\displaystyle p=1}
(G ) becomes the harmonic series which we saw was divergent, and neither of the above tests apply.
When
p
<
1
{\displaystyle p<1}
(G ) will, after the first, be greater than the corresponding terms of the harmonic series; hence (G ) is divergent.
139. Cauchy's ratio test for convergence.
(A )
u
1
+
u
2
+
u
3
+
⋯
{\displaystyle u_{1}+u_{2}+u_{3}+\cdots }
be a series of positive termes to be tested.
