Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/240

137.Fundamental test for convergence. Summing up first $n$ and then $n+p$ terms of a series, we have

(A)	$S_{n}=u_{1}+u_{2}+u_{3},+\cdots +u_{n}$,

(B)	$S_{n+p}=u_{1}+u_{2}+u_{3},+\cdots +u_{n}+u_{n+1}+\cdots +u_{n+p}$,

Subtracting (A) from (B),

(C)	$S_{n+p}-S_{n}=u_{n+1}+u_{n_{2}}+\cdots +u_{n+p}$.

From Theorem III we know that the necessary and sufficient condition that the series shall be convergent is that

$\lim _{n\to \infty }\left(S_{n+p}-S_{n}\right)=0$

for every value of $p$. But this is the same as the left-hand member of (C); therefore from the right-hand member the condition may also be written

(D)	$\lim _{n\to \infty }\left(u_{n+1}+u_{n+2}+\cdots +u_{n+p}\right)=0$.

Since (D) is true for every value of $p$, then, letting $p=1$, a necessary condition for convergence is that

$\lim _{n\to \infty }\left(u_{n+1}\right)=0$;

or, what amounts to the same thing,

(E)	$\lim _{n\to \infty }\left(u_{n}\right)=0$.

Hence, if the general (or nth) term of a series does not approach zero as $n$ approaches infinity, we know at once that the series is non- convergent and we need proceed no further. However, (D) is not a sufficient condition; that is, even if the nth term does approach zero, we cannot state positively that the series is convergent ; for, consider the harmonic series

$1,\,{\frac {1}{2}},\,{\frac {1}{3}},\,{\frac {1}{4}},\,\cdots ,\,{\frac {1}{n}}$.

Here

$\lim _{n\to \infty }\left(u_{n}\right)=\lim _{n\to \infty }\left({\frac {1}{n}}\right)=0$;

that is, condition (E) is fulfilled. Yet we may show that the harmonic series is not convergent by the following comparison:

(F)	$1+{\frac {1}{2}}+\left[{\frac {1}{3}}+{\frac {1}{4}}\right]+\left[{\frac {1}{5}}+{\frac {1}{6}}+{\frac {1}{7}}+{\frac {1}{8}}\right]+\left[{\frac {1}{9}}+\cdots +{\frac {1}{16}}\right]+\cdots$,

(G)	${\frac {1}{2}}+{\frac {1}{2}}+\left[{\frac {1}{4}}+{\frac {1}{4}}\right]+\left[{\frac {1}{8}}+{\frac {1}{8}}+{\frac {1}{8}}+{\frac {1}{8}}\right]+\left[{\frac {1}{16}}+\cdots +{\frac {1}{16}}\right]+\cdots$,