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D’ ALEMBERT’s Test of Convergence of Series

Saturday, April 9th, 2011 11:36 / 1 Comment

Statement

A series $\sum {u_n}$ of positive terms is convergent if from and after some fixed term $\dfrac {u_{n+1}} {u_n} < r < {1}$ , where r is a fixed number. The series is divergent if $\dfrac{u_{n+1}} {u_n} > 1$ from and after some fixed term.

D’ Alembert’s Test is also known as ratio test of convergency of a series.

Definitions for Generally Interested Readers

(Definition 1) An infinite series $\sum {u_n}$ i.e. $\mathbf {u_1+u_2+u_3+....+u_n}$ is said to be convergent if $S_n$ , the sum of its first $n$ terms, tends to a finite limit $S$ as n tends to infinity.
We call $S$ the sum of the series, and write $S=\displaystyle {\lim_{n \to \infty} } S_n$ .
Thus an infinite series $\sum {u_n}$ converges to a sum S, if for any given positive number $\epsilon$ , however small, there exists a positive integer $n_0$ such that
$|S_n-S| < \epsilon$ for all $n \ge n_0$ .
(Definition 2)
If $S_n \to \pm \infty$ as $n \to \infty$ , the series is said to be divergent.
Thus, $\sum {u_n}$ is said to be divergent if for every given positive number $\lambda$ , however large, there exists a positive integer $n_0$ such that $|S_n|>\lambda$ for all $n \ge n_0$ .
(Definition 3)
If $S_n$ does not tends to a finite limit, or to plus or minus infinity, the series is called Oscillatory

Discussions

Let a series be $\mathbf {u_1+u_2+u_3+.......}$ . We assume that the above inequalities are true.

From the first part of the statement:
$\dfrac {u_2}{u_1} < r$ , $\dfrac {u_3}{u_2} < r$ ……… where r <1.
Therefore $\mathbf {{u_1+u_2+u_3+....}= u_1 {(1+\frac{u_2}{u_1}+\frac{u_3}{u_1}+....)}}$
$=\mathbf {u_1{(1+\frac{u_2}{u_1}+\frac{u_3}{u_2} \times \frac{u_2}{u_1}+....)}}$
$< \mathbf {u_1(1+r+r^2+.....)}$
Therefore, $\sum{u_n} < u_1 (1+r+r^2+.....)$
or, $\sum{u_n} < \displaystyle{\lim_{n \to \infty}} \dfrac {u_1 (1-r^n)} {1-r}$
Since r<1, therefore as $n \to \infty , \ r^n \to 0$
therefore $\sum{u_n} < \dfrac{u_1} {1-r}$ =k say, where k is a fixed number.
Therefore $\sum{u_n}$ is convergent.
Since, $\dfrac{u_{n+1}}{u_n} > 1$ then, $\dfrac{u_2}{u_1} > 1$ , $\dfrac{u_3}{u_2} > 1$ …….
Therefore $u_2 > u_1, \ u_3 >u_2>u_1, \ u_4 >u_3 > u_2 >u_1$ and so on.
Therefore $\sum {u_n}=u_1+u_2+u_3+....+u_n$ > $nu_1$ . By taking n sufficiently large, we see that $nu_1$ can be made greater than any fixed quantity.
Hence the series is divergent.

Comments

When $\dfrac {u_{n+1}} {u_n}=1$ , the test fails.

Another form of the test–

The series $\sum {u_n}$ of positive terms is convergent if $\displaystyle {\lim_{n \to \infty}} \dfrac {u_n}{u_{n+1}}$ >1 and divergent if $\displaystyle{\lim_{n \to \infty}} \dfrac {u_n}{u_{n+1}}$ <1.
One should use this form of the test in the practical applications.

A Problem:
Verify whether the infinite series $\dfrac{x}{1.2} + \dfrac {x^2} {2.3} + \dfrac {x^3} {3.4} +....$ is convergent or divergent.

Solution

We have $u_{n+1}= \dfrac {x^{n+1}}{(n+1)(n+2)}$ and $u_n= \dfrac {x^n} {n(n+1)}$
Therefore $\displaystyle {\lim_{n \to \infty}} \dfrac{u_n} {u_{n+1}} = \displaystyle{\lim_{n \to \infty}} (1+\frac{2}{n}) \frac{1}{x} = \frac{1}{x}$
Hence, when 1/x >1 , i.e., x <1, the series is convergent and when x >1 the series is divergent.
When x=1, $u_n=\dfrac{1} {n(n+1)}=\dfrac {1}{n^2} {(1+1/n)}^{-1}$
or, $u_n=\dfrac{1}{n^2}(1-\frac{1}{n}+ \frac {1}{n^2}-.....)$
Take $\dfrac{1}{n^2}=v_n$ Now $\displaystyle {\lim_{n \to \infty}} \dfrac {u_n}{v_n}=1$ , a non-zero finite quantity.
But $\sum {v_n}=\sum {\frac{1}{n^2}}$ is convergent.
Hence, $\sum {u_n}$ is also Convergent.