# 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 $ \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.

**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.

## Suggested Reading

- Hardcover Book
- Tao, Terence (Author)
- English (Publication Language)

## 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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You can either start a new conversation or continue an existing one.Please don't use this comment form just to build backlinks. If your comment is not good enough and if in some ways you are trying to just build links — your comment will be deleted. Use this form to build a better and cleaner commenting ecosystem. Students are welcome to ask for help, freebies and more. Your email will not be published or used for any purposes.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??

When the test fails…you have to use raabes test….

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

Thanks for the brilliant explanation . u enjoyed and understood it better than I got in the lecture

Unable to understand