Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/248

We may then write down the following

General directions for finding the interval of convergence of the power series,

(A)	$a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+\cdots .$

First Step. Write down the series formed by coefficients, namely,

$a_{0}+a_{1}+a_{2}+a_{3}+\cdots +a_{n}+a_{n+1}+\cdots .$

Second Step. Calculate the limit

$\lim _{n\to \infty }\left({\frac {a_{n+1}}{a_{n}}}\right)=L.$

Third Step.. Then the power series (A) is

I.	Absolutely convergent for all values of $x$ lying between.
$-|{\tfrac {1}{L}}|$ and $+|{\tfrac {1}{L}}|.$
II.	Divergent for all values of x less than $-|{\tfrac {1}{L}}|$ or greater than $-|{\tfrac {1}{L}}|$.
III.	No test when $x=\pm |{\tfrac {1}{L}}|$; but then we substitute these two values of $x$ in the power series (A) and apply to them the general directions in §141.

Note. When $L=0,\pm |{\tfrac {1}{L}}|=\pm \infty$ and the power series is absolutely convergent for all values of $x$.

Illustrative Example 1. Find the interval of convergence for the series

(B)	$x-{\frac {x^{2}}{2^{2}}}+{\frac {x^{3}}{3^{2}}}-{\frac {x^{4}}{4^{2}}}+\cdots .$

Solution. First step. The series formed by the coefficients is

(C)	$x-{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}-{\frac {1}{4^{2}}}+\cdots .$

Second step.	$\lim _{n\to \infty }\left({\frac {a_{n+1}}{a_{n}}}\right)=$	$\lim _{n\to \infty }\left[-{\frac {n^{2}}{\left(n+1\right)^{2}}}\right]={\frac {\infty }{\infty }}.$
Differentiating,		$\lim _{n\to \infty }\left(-{\frac {2n}{2\left(n+1\right)}}\right)={\frac {\infty }{\infty }}.$
Differentiating again,		$\lim _{n\to \infty }\left(-{\frac {2}{2}}\right)=-1\left(L\right).$
Third Step	$\left|{\frac {1}{L}}\right|=\left|{\frac {1}{-1}}\right|=1.$

By I the series is absolutely convergent when $x$ lies between $-1$ and $+1$.

By II the series is divergent when $x$ is less than $1$ or greater than $+1$.

By III there is no test when $x=\pm 1$.