D’Alembert’s Ratio Test of Convergence of Series

In this article we will formulate the D’ Alembert’s Ratio Test on convergence of a series.

Let’s start.

Statement of D’Alembert Ratio Test

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 the ratio test of convergence of a series.

Theorem

Let $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ be a series of real numbers in $R$, or a series of complex numbers in $C$.

Let the sequence ${a_n}$ satisfy:
$\displaystyle \lim_{n \mathop \to \infty} {\frac {a_{n + 1} } {a_n} } = l$

  • If $l > 1 $, then $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ diverges.
  • If $l < 1 $, then $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ converges absolutely.

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 tend to a finite limit, or to plus or minus infinity, the series is called oscillatory.

Proof & Discussions on Ratio Test

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 $$ {u_1+u_2+u_3+….} \\ = u_1 {(1+\frac{u_2}{u_1}+\frac{u_3}{u_1}+….)}$$
$ =u_1{(1+\frac{u_2}{u_1}+\frac{u_3}{u_2} \times \frac{u_2}{u_1}+….)} $
$ < 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.

Academic Proof

From the statement of the theorem, it is necessary that $\forall n: a_n \ne 0$; otherwise ${\dfrac {a_{n + 1} } {a_n} }$ is not defined.
Here, ${\dfrac {a_{n + 1} } {a_n} }$ denotes either the absolute value of $\dfrac {a_{n + 1} } {a_n}$, or the complex modulus of $\dfrac {a_{n + 1} } {a_n}$.

Absolute Convergence
Suppose $l < 1$.
Let us take $\epsilon > 0$ such that $l + \epsilon < 1$.
Then:
$\exists N: \forall n > N: {\dfrac {a_n} {a_{n – 1} } } < l + \epsilon$
Thus:
$ (\displaystyle {a_n}) = (\displaystyle {\frac {a_n} {a_{n – 1} } } {\frac {a_{n – 1} } {a_{n – 2} } } \dotsm {\frac {a_{N + 2} } {a_{N + 1} } } {a_{N + 1} }) \\ < (\displaystyle {l + \epsilon}^{n – N – 1} {a_{N + 1} })$
By Sum of Infinite Geometric Progression, $\displaystyle \sum_{n \mathop = 1}^\infty{l + \epsilon}^n$ converges.
So by the the corollary to the comparison test, it follows that $\displaystyle \sum_{n \mathop = 1}^\infty {a_n}$ converges absolutely too.
$\blacksquare$

Divergence
Suppose $l > 1$.
Let us take $\epsilon > 0$ small enough that $l – \epsilon > 1$.
Then, for a sufficiently large $N$, we have:
$(\displaystyle {a_n}) = (\displaystyle {\frac {a_n} {a_{n – 1} } } {\frac {a_{n – 1} }{a_{n – 2} } } \dotsm {\frac {a_{N + 2} } {a_{N + 1} } } {a_{N + 1} })$
$> (\displaystyle {l – \epsilon}^{n – N + 1} {a_{N + 1} })$
But ${l – \epsilon}^{n – N + 1} {a_{N + 1} } \to \infty$ as $n \to \infty$.
So $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ diverges.
$\blacksquare$

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.

Suggested Reading

An Example

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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Last update on 2022-06-22 using Amazon Product Advertising API

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