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# D’Alembert's Ratio Test (With Solved Exercise) Mathematics Notes | EduRev

## Mathematics : D’Alembert's Ratio Test (With Solved Exercise) Mathematics Notes | EduRev

``` Page 1

Article D’Alembert

s Ratio Test
Statement If ?u

is a series of positive terms such that
(a)  lim

u

u

then
(i)     ?u

is convergent if                            (ii) ?u

is divergent if
(iii)  ?u

may converge or diverge if     (i e ,the test fails if    )
(b)   lim

u

u

,then the series ?u

is convergent
Proof (a) ?u

is series of positive terms
u

n

u

u

lim

u

u

Since    lim

u

u

,therefore,for each       a positive integer m such that
|
u

u

|               n  m

u

u

Replacing n by m,m  ,m  ,  ,n    in the above inequality,we have

u

u

u

u

u

u

u

u

Multiplying the above (n m) inequality,we have
(   )

u

u

(   )

( )
(i) Let
Choose     such that      (This is always possible since we have only to choose an   such that

From ( ),we have
(   )

u

u

u

u

(   )

u

u

(   )

(   )

u

K

(   )

n m           where K u

(   )

Now,?

(   )

being a geometric series with common ratio

is convergent Therefore,
Page 2

Article D’Alembert

s Ratio Test
Statement If ?u

is a series of positive terms such that
(a)  lim

u

u

then
(i)     ?u

is convergent if                            (ii) ?u

is divergent if
(iii)  ?u

may converge or diverge if     (i e ,the test fails if    )
(b)   lim

u

u

,then the series ?u

is convergent
Proof (a) ?u

is series of positive terms
u

n

u

u

lim

u

u

Since    lim

u

u

,therefore,for each       a positive integer m such that
|
u

u

|               n  m

u

u

Replacing n by m,m  ,m  ,  ,n    in the above inequality,we have

u

u

u

u

u

u

u

u

Multiplying the above (n m) inequality,we have
(   )

u

u

(   )

( )
(i) Let
Choose     such that      (This is always possible since we have only to choose an   such that

From ( ),we have
(   )

u

u

u

u

(   )

u

u

(   )

(   )

u

K

(   )

n m           where K u

(   )

Now,?

(   )

being a geometric series with common ratio

is convergent Therefore,
For more notes,  call 8130648819

by comparison test,the series ? u

is convergent
(ii) Let
Choose     such that       (This is always possiblle since we have only to choose an   such that

Since    ,
From ( ),we have

u

u

(   )

u

u

(   )

u

u

(   )

(   )

u

(   )

n m where   u

(   )

Now,?

(   )

being a geometric series with common ratio

is divergent Therefore,
by comparison test,the series ? u

is divergent
(iii) Let
First consider the series
?

n

n

u

n
,                              u

n

u

u

n
n

n

so that
lim

u

u

Since the series ?

n
is divergent,we find that if    ,a series may diverge
Next,consider the series
?

n

n

u

n

,                          u

(n  )

u

u

(n  )

n

(
n
n
)

(

n
)

so that Lim

u

u

Since the series ?

n

is convergent,we find that if
,a series may converge The above two examples show that if    ,a series may converge or diverge
Hence the test fails when
Remar  Another equivalent form of Ratio Test is as follow
If ?u

is a positive term series such that lim

u

u

then
(i) ?u

is convergent if
(ii) ?u

is divergent if

Ex    Discuss the convergence of the following series
(i)

(ii)

,(p  )
(iii)

(iv)

Solution (i) Here u

n
n

u

(n  )
(n  )

Page 3

Article D’Alembert

s Ratio Test
Statement If ?u

is a series of positive terms such that
(a)  lim

u

u

then
(i)     ?u

is convergent if                            (ii) ?u

is divergent if
(iii)  ?u

may converge or diverge if     (i e ,the test fails if    )
(b)   lim

u

u

,then the series ?u

is convergent
Proof (a) ?u

is series of positive terms
u

n

u

u

lim

u

u

Since    lim

u

u

,therefore,for each       a positive integer m such that
|
u

u

|               n  m

u

u

Replacing n by m,m  ,m  ,  ,n    in the above inequality,we have

u

u

u

u

u

u

u

u

Multiplying the above (n m) inequality,we have
(   )

u

u

(   )

( )
(i) Let
Choose     such that      (This is always possible since we have only to choose an   such that

From ( ),we have
(   )

u

u

u

u

(   )

u

u

(   )

(   )

u

K

(   )

n m           where K u

(   )

Now,?

(   )

being a geometric series with common ratio

is convergent Therefore,
For more notes,  call 8130648819

by comparison test,the series ? u

is convergent
(ii) Let
Choose     such that       (This is always possiblle since we have only to choose an   such that

Since    ,
From ( ),we have

u

u

(   )

u

u

(   )

u

u

(   )

(   )

u

(   )

n m where   u

(   )

Now,?

(   )

being a geometric series with common ratio

is divergent Therefore,
by comparison test,the series ? u

is divergent
(iii) Let
First consider the series
?

n

n

u

n
,                              u

n

u

u

n
n

n

so that
lim

u

u

Since the series ?

n
is divergent,we find that if    ,a series may diverge
Next,consider the series
?

n

n

u

n

,                          u

(n  )

u

u

(n  )

n

(
n
n
)

(

n
)

so that Lim

u

u

Since the series ?

n

is convergent,we find that if
,a series may converge The above two examples show that if    ,a series may converge or diverge
Hence the test fails when
Remar  Another equivalent form of Ratio Test is as follow
If ?u

is a positive term series such that lim

u

u

then
(i) ?u

is convergent if
(ii) ?u

is divergent if

Ex    Discuss the convergence of the following series
(i)

(ii)

,(p  )
(iii)

(iv)

Solution (i) Here u

n
n

u

(n  )
(n  )

For more notes,  call 8130648819

u

u

n
n

(n  )

(n  )

n
n

(n  )

(n  )n

(n  )

n

(
n
n
)

(

n
)

lim

u

u

lim

(

n
)

e                            ,   e  -
y D’Alembert’s Ratio Test, u

is convergent
(ii) Do yourself                                                       ,Ans Convergent-
(iii) Do yourself                                                      ,Ans Convergent-
(iv) Do yourself                                                      ,Ans Convergent-
JAM      MCQ Que
Let S be the series ?

(    )
(    )

and T be the series ? 4

5
(   )

of real numbers
Then,which one of the following is true
(a)  oth the series S and T are convergent
(b) S is convergent and T is divergent
(c) S is divergent and T is convergent
(d)  oth the series S and T are divergent
Sol

S ?

(    )
(    )

?

convergent
or Consider the u

for S
u

( n  )
(    )

u

, (n  )  -
, (   )  -

, n  -,

-

u

u

( n  )(

)
( n  )(

)

n.

n
/(

)
n.

n
/(

)

.

n
/
.

n
/

lim

u

u

by ratio test S is convergent
Consider the u

for T
u

(
n
n
)
(   )

u

(
n
n
)
(   )

u

u

(
n
n
)

(
n
n
)

:

n

n
;

:

n

n
;

lim

u

u

tends to
So, T is divergent
Ex    Test the convergence of the following series
(i)

(ii)

(iii)

(iv)

(v)

Solution (i) Here u

(n)

u

(n  )

u

u

(n  )

n

n

n
n

n

n

n

Page 4

Article D’Alembert

s Ratio Test
Statement If ?u

is a series of positive terms such that
(a)  lim

u

u

then
(i)     ?u

is convergent if                            (ii) ?u

is divergent if
(iii)  ?u

may converge or diverge if     (i e ,the test fails if    )
(b)   lim

u

u

,then the series ?u

is convergent
Proof (a) ?u

is series of positive terms
u

n

u

u

lim

u

u

Since    lim

u

u

,therefore,for each       a positive integer m such that
|
u

u

|               n  m

u

u

Replacing n by m,m  ,m  ,  ,n    in the above inequality,we have

u

u

u

u

u

u

u

u

Multiplying the above (n m) inequality,we have
(   )

u

u

(   )

( )
(i) Let
Choose     such that      (This is always possible since we have only to choose an   such that

From ( ),we have
(   )

u

u

u

u

(   )

u

u

(   )

(   )

u

K

(   )

n m           where K u

(   )

Now,?

(   )

being a geometric series with common ratio

is convergent Therefore,
For more notes,  call 8130648819

by comparison test,the series ? u

is convergent
(ii) Let
Choose     such that       (This is always possiblle since we have only to choose an   such that

Since    ,
From ( ),we have

u

u

(   )

u

u

(   )

u

u

(   )

(   )

u

(   )

n m where   u

(   )

Now,?

(   )

being a geometric series with common ratio

is divergent Therefore,
by comparison test,the series ? u

is divergent
(iii) Let
First consider the series
?

n

n

u

n
,                              u

n

u

u

n
n

n

so that
lim

u

u

Since the series ?

n
is divergent,we find that if    ,a series may diverge
Next,consider the series
?

n

n

u

n

,                          u

(n  )

u

u

(n  )

n

(
n
n
)

(

n
)

so that Lim

u

u

Since the series ?

n

is convergent,we find that if
,a series may converge The above two examples show that if    ,a series may converge or diverge
Hence the test fails when
Remar  Another equivalent form of Ratio Test is as follow
If ?u

is a positive term series such that lim

u

u

then
(i) ?u

is convergent if
(ii) ?u

is divergent if

Ex    Discuss the convergence of the following series
(i)

(ii)

,(p  )
(iii)

(iv)

Solution (i) Here u

n
n

u

(n  )
(n  )

For more notes,  call 8130648819

u

u

n
n

(n  )

(n  )

n
n

(n  )

(n  )n

(n  )

n

(
n
n
)

(

n
)

lim

u

u

lim

(

n
)

e                            ,   e  -
y D’Alembert’s Ratio Test, u

is convergent
(ii) Do yourself                                                       ,Ans Convergent-
(iii) Do yourself                                                      ,Ans Convergent-
(iv) Do yourself                                                      ,Ans Convergent-
JAM      MCQ Que
Let S be the series ?

(    )
(    )

and T be the series ? 4

5
(   )

of real numbers
Then,which one of the following is true
(a)  oth the series S and T are convergent
(b) S is convergent and T is divergent
(c) S is divergent and T is convergent
(d)  oth the series S and T are divergent
Sol

S ?

(    )
(    )

?

convergent
or Consider the u

for S
u

( n  )
(    )

u

, (n  )  -
, (   )  -

, n  -,

-

u

u

( n  )(

)
( n  )(

)

n.

n
/(

)
n.

n
/(

)

.

n
/
.

n
/

lim

u

u

by ratio test S is convergent
Consider the u

for T
u

(
n
n
)
(   )

u

(
n
n
)
(   )

u

u

(
n
n
)

(
n
n
)

:

n

n
;

:

n

n
;

lim

u

u

tends to
So, T is divergent
Ex    Test the convergence of the following series
(i)

(ii)

(iii)

(iv)

(v)

Solution (i) Here u

(n)

u

(n  )

u

u

(n  )

n

n

n
n

n

n

n

For more notes,  call 8130648819

lim

u

u

lim

n

n

n

y D

Alembert

s Ratio Test,?u

is divergent
(ii) The given series is

Here u

n

u

(n  )

u

u

n

(n  )

n

n

.

n
/

.

n
/

lim

u

u

lim

.

n
/

y D

Alembert

s Ratio Test, u

s convergent
(iii) Do yourself                                                          ,Ans Divergent-
(iv) u

n

, u

(n  )

divergent

u

u

n

(n  )

n

lim

u

u

y D

Alembert Ratio Test,the series is divergent
(v) Do yourself                                               ,Ans Divergent-
Ex    Examine the convergence or divergence of the following series
(i)

(ii)

(iii) ?

Solution (i) Here
u

n
,  (n  )  -

n
( n  )

u

n(n  )
( n  )( n  )

u

u

n
( n  )
( n  )( n  )
n(n  )

u

u

( n  )
n

n

n

lim

u

u

lim

n

n

y D

Alembert’s Ratio  Test, u

is convergent
(ii) Here u

u

u

u

.

/

.

/

lim

u

u

lim

y D

Alembert’s Ratio Test, u

is divergent
(iii) Do yourself                                                        ,Ans Convergent-
Ex    Test the convergence of the following series
(i) ?

n
(ii) ?
n

a

a
(iii) ?
n

(n  )

n
(iv) ?

n
n

Solution (i) Do yourself                                                                      ,Ans Convergent-
Page 5

Article D’Alembert

s Ratio Test
Statement If ?u

is a series of positive terms such that
(a)  lim

u

u

then
(i)     ?u

is convergent if                            (ii) ?u

is divergent if
(iii)  ?u

may converge or diverge if     (i e ,the test fails if    )
(b)   lim

u

u

,then the series ?u

is convergent
Proof (a) ?u

is series of positive terms
u

n

u

u

lim

u

u

Since    lim

u

u

,therefore,for each       a positive integer m such that
|
u

u

|               n  m

u

u

Replacing n by m,m  ,m  ,  ,n    in the above inequality,we have

u

u

u

u

u

u

u

u

Multiplying the above (n m) inequality,we have
(   )

u

u

(   )

( )
(i) Let
Choose     such that      (This is always possible since we have only to choose an   such that

From ( ),we have
(   )

u

u

u

u

(   )

u

u

(   )

(   )

u

K

(   )

n m           where K u

(   )

Now,?

(   )

being a geometric series with common ratio

is convergent Therefore,
For more notes,  call 8130648819

by comparison test,the series ? u

is convergent
(ii) Let
Choose     such that       (This is always possiblle since we have only to choose an   such that

Since    ,
From ( ),we have

u

u

(   )

u

u

(   )

u

u

(   )

(   )

u

(   )

n m where   u

(   )

Now,?

(   )

being a geometric series with common ratio

is divergent Therefore,
by comparison test,the series ? u

is divergent
(iii) Let
First consider the series
?

n

n

u

n
,                              u

n

u

u

n
n

n

so that
lim

u

u

Since the series ?

n
is divergent,we find that if    ,a series may diverge
Next,consider the series
?

n

n

u

n

,                          u

(n  )

u

u

(n  )

n

(
n
n
)

(

n
)

so that Lim

u

u

Since the series ?

n

is convergent,we find that if
,a series may converge The above two examples show that if    ,a series may converge or diverge
Hence the test fails when
Remar  Another equivalent form of Ratio Test is as follow
If ?u

is a positive term series such that lim

u

u

then
(i) ?u

is convergent if
(ii) ?u

is divergent if

Ex    Discuss the convergence of the following series
(i)

(ii)

,(p  )
(iii)

(iv)

Solution (i) Here u

n
n

u

(n  )
(n  )

For more notes,  call 8130648819

u

u

n
n

(n  )

(n  )

n
n

(n  )

(n  )n

(n  )

n

(
n
n
)

(

n
)

lim

u

u

lim

(

n
)

e                            ,   e  -
y D’Alembert’s Ratio Test, u

is convergent
(ii) Do yourself                                                       ,Ans Convergent-
(iii) Do yourself                                                      ,Ans Convergent-
(iv) Do yourself                                                      ,Ans Convergent-
JAM      MCQ Que
Let S be the series ?

(    )
(    )

and T be the series ? 4

5
(   )

of real numbers
Then,which one of the following is true
(a)  oth the series S and T are convergent
(b) S is convergent and T is divergent
(c) S is divergent and T is convergent
(d)  oth the series S and T are divergent
Sol

S ?

(    )
(    )

?

convergent
or Consider the u

for S
u

( n  )
(    )

u

, (n  )  -
, (   )  -

, n  -,

-

u

u

( n  )(

)
( n  )(

)

n.

n
/(

)
n.

n
/(

)

.

n
/
.

n
/

lim

u

u

by ratio test S is convergent
Consider the u

for T
u

(
n
n
)
(   )

u

(
n
n
)
(   )

u

u

(
n
n
)

(
n
n
)

:

n

n
;

:

n

n
;

lim

u

u

tends to
So, T is divergent
Ex    Test the convergence of the following series
(i)

(ii)

(iii)

(iv)

(v)

Solution (i) Here u

(n)

u

(n  )

u

u

(n  )

n

n

n
n

n

n

n

For more notes,  call 8130648819

lim

u

u

lim

n

n

n

y D

Alembert

s Ratio Test,?u

is divergent
(ii) The given series is

Here u

n

u

(n  )

u

u

n

(n  )

n

n

.

n
/

.

n
/

lim

u

u

lim

.

n
/

y D

Alembert

s Ratio Test, u

s convergent
(iii) Do yourself                                                          ,Ans Divergent-
(iv) u

n

, u

(n  )

divergent

u

u

n

(n  )

n

lim

u

u

y D

Alembert Ratio Test,the series is divergent
(v) Do yourself                                               ,Ans Divergent-
Ex    Examine the convergence or divergence of the following series
(i)

(ii)

(iii) ?

Solution (i) Here
u

n
,  (n  )  -

n
( n  )

u

n(n  )
( n  )( n  )

u

u

n
( n  )
( n  )( n  )
n(n  )

u

u

( n  )
n

n

n

lim

u

u

lim

n

n

y D

Alembert’s Ratio  Test, u

is convergent
(ii) Here u

u

u

u

.

/

.

/

lim

u

u

lim

y D

Alembert’s Ratio Test, u

is divergent
(iii) Do yourself                                                        ,Ans Convergent-
Ex    Test the convergence of the following series
(i) ?

n
(ii) ?
n

a

a
(iii) ?
n

(n  )

n
(iv) ?

n
n

Solution (i) Do yourself                                                                      ,Ans Convergent-
For more notes,  call 8130648819

(ii) Here u

n

a

a
u

(n  )

a

a

u

u

(n

a)(

a)
((n  )

a)(

a)

n

.
a
n

/
n

6.

n
/

a
n

7

.
a

/

.
a

/

a
n

.

n
/

a
n

.
a

/

a

lim

u

u

y D

Alembert

s Ratio Test, u

is convergent
(iii) Do yourself                                                                          ,Ans Convergent-
(iv) Do yourself                                                                           ,Ans Convergent-
Ex    Discuss the convergence or divergence of the following series
(i) ?
n

n
(ii) ?
n
n

(iii) ?

n
n

Solution (i) Here u

n

n

u

(n  )

(n  )

(n  )

(n  )n

n
n

u

u

n

n

n

n

lim

u

u

lim

n

n

y D

Alembert

s Ratio Test, u

is convergent
(ii) Do yourself                                                                        ,Ans Convergent-
(iii) Do yourself                                                                       ,Ans Divergent-
Ex    Discuss the convergence or divergence of the following series
(i) ?
x

n

,x                (ii) ?
x

a vn
(iii) ?
v
n
n

x

(iv) ?
vn
vn

x

(v) ?
x

n
,x                     (vi) ?
n
n

x

,x
Solution (i) Here u

x

n

u

x

(n  )

u

u

x

x

(n  )

n

x
(

n
)

lim

u

u

lim

x
(

n
)

x

y D

Alembert

s Ratio Test, u

converges if

x
i e x   and diverges if

x
i e ,x
When x  ,
lim

u

u

Ratio Test Fails
Now,when x  ,
u

n

n

?u

?

n

is of the form ?

n

with p
?u

is convergent
Hence the given series  u

converges if x   and diverges if x
(ii) Do yourself                               ,Ans converges if x   and diverges if x  -
```
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## Topic-wise Tests & Solved Examples for IIT JAM Mathematics

27 docs|150 tests

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