D’ ALEMBERT’s Test of Convergence of Series

D’ Alembert’s Test is also known as ratio test of convergence 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.