Home » Math » D’ ALEMBERT’s Test of Convergence of Series

D’ ALEMBERT’s Test of Convergence of Series

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.

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1 Comment

  1. WordPress URL Shortening: An Interesting Consequence | MY DIGITAL NOTEBOOK says:
    Thursday, November 24th, 2011 at 12:05

    [...] Where does it go? It goes to d’ Alembert’s Test of Convergence . Reason is specified below at point [...]

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