## StatementA 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 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.

## 11 comments

I love this digital notebook. It is very helpful to me. Thanks

Thank you for your kind words.

Thanks for you support in other to understand this part of D’Alembert’s ratio test. Can you give more examples for understand it more better? thanks.

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

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