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 $\sequence {a_n}$ 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_nS < \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 $ \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 (1r^n)} {1r}$
Since r<1, therefore as $ n \to \infty , \ r^n \to 0$
therefore $ \sum{u_n} < \dfrac{u_1} {1r}$ =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} })$
$(\displaystyle ) (<) (\displaystyle {l + \epsilon}^{n – N – 1} {a_{N + 1} })$
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:
$(\displaystyle {a_n}) (=) (\displaystyle {\frac {a_n} {a_{n – 1} } } {\frac {a_{n – 1} }{a_{n – 2} } } \dotsm {\frac {a_{N + 2} } {a_{N + 1} } } {a_{N + 1} })$
$(\displaystyle ) (>) (\displaystyle {l – \epsilon}^{n – N + 1} {a_{N + 1} })$
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.
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 nonzero finite quantity.
But $ \sum {v_n}=\sum {\frac{1}{n^2}}$ is convergent.
Hence, $ \sum {u_n}$ is also convergent.
16 comments add your comment
here is D’ Alembert’s ratio test:
Let Un be the nth term of a positive series such that
lim Un+1/Un = L
Then the series is convergent if L 1.
The test fails to decide the nature of the series if L = 1.
series is convergent if L 1
When we get the result 1′ we have to work this out with other principle…what did you do when you get the result equal to 1??
here is D’ Alembert’s ratio test:
Let Un be the nth term of a positive series such that
lim Un+1/Un = L
Then the series is convergent if L 1.
The test fails to decide the nature of the series if L = 1.
series is convergent if L 1
When we get the result 1′ we have to work this out with other principle…what did you do when you get the result equal to 1??
Series is convergent if L is less than 1 and divergent if L is greater than 1
Series is convergent if L is less than 1 and divergent if L is greater than 1
It’s nice. Ratio test has so many forms due to which creates confusion. I applied ratio test in this series
1+ (1/2!)+ (1/3!)+…..
But I found this series to be divergent using ratio test while this series is convergent.
i hate the question having factorial of something in the denominator
It’s nice. Ratio test has so many forms due to which creates confusion. I applied ratio test in this series
1+ (1/2!)+ (1/3!)+…..
But I found this series to be divergent using ratio test while this series is convergent.
i hate the question having factorial of something in the denominator
Thanks for the brilliant explanation . u enjoyed and understood it better than I got in the lecture
Thanks for the brilliant explanation . u enjoyed and understood it better than I got in the lecture
Unable to understand
Unable to understand