infinite series.docx

INFINITE SERIES

Sequence:

If a set of real numbers u1 ,u2 ,……,un occur according to some definite rule, then it is called

a sequence denoted by {Sn }={u1 , u2 ,……,un } if n is finite

Or {Sn }={u1 , u2 ,……,un ,…….} if n is infinite.

Series:

u1+u2+……+un is called a series and is denoted by Sn=∑k=1

n

uk

Infinite Series:

If the number of terms in the series is infinitely large, then it is called infinite series and is denoted by ∑ un=¿u1+u2+……+un+……¿ and the sum of its first n terms be denoted by

Sn=∑k=1

n

uk=u1+u2+……+un.

Convergence:

An infinite series ∑ un is said to be convergent if limn→∞

Sn=k, a definite unique number.

Example: 1+ 12+ 1

4+…… ..

Sn=1+ 12+ 1

22 +…… ..+ 12n−1=

(1−1

2n )(1−1

2 )=2, finite.

Therefore given series is convergent.

Divergence:

limn→∞

Sn tends to either ∞∨−∞ then the infinite series ∑ un is said to be divergent.

Example: ∑ un=1+2+3+……. .

Sn=n (n+1 )

2

limn→∞

Sn=∞

Therefore ∑ un is divergent.

Oscillatory Series:

If limn→∞

Sn tends to more than one limit either finite or infinite, then the infinite series ∑ un is

said to be oscillatory series.

Example:1. ∑ un=1−1+1−1+…… ..:

Sn={1 ,n is odd0 , n is even

Therefore series is oscillatory.

2. ∑ un=1+(−3)+(−3)2+…… ..

Sn=1−(−1)n3n

1+3

limn→∞

Sn={ ∞,n is odd−∞ ,n is even

Properties of infinite series:

1. The convergence or divergence of an infinite series remains unaltered on multiplication of each term by c ≠o.

2. The convergence or divergence of an infinite series remains unaltered by addition or removal of a finite number of its terms.

Positive term series:

An infinite series in which all the terms after some particular term are positive is called a positive term series.

Geometric Series test:

The series ∑n=0

∞

rn=1+r+r 2+r 3+……+rn+……

a. Converges if |r|<1b. Diverges if r ≥1c. Oscillates finitely if r=−1 and oscillates infinitely if r←1

Proof:

Let Sn be the partial sum of ∑n=0

∞

rn.

Sn=1+r+r2+…+r n−1

Case 1:|r|<1 i.e. −1<r<1

Sn=1−rn

1−r

limn→∞

Sn=1

1−r

Therefore the series is convergent.

Case 2i:r>1 i.e. limn→∞

rn=∞

Sn=rn−1r−1

limn→∞

Sn=∞

Therefore the series is divergent.

Case 2ii: r=1,Sn=1+1+1+1+…….+1=n limn→∞

Sn=∞.

Therefore the series is divergent.

Case 3i:r←1 i.e. Let r=−m

Sn=1−rn

1−r=

1−(−1)nmn

1+m

limn→∞

Sn={ ∞,n is odd−∞ ,n is even

Therefore the series is oscillatory.

Case 3ii:r=−1

i.e. Sn=1−1+1−1+…….

limn→∞

Sn={1 , n is odd0 , n is even

Therefore the series is oscillatory.

Note: If a series in which all the terms are positive is convergent, the series remains convergent even when some or all of its terms are negative.

Integral Test:

A positive term series f (1 )+ f (2 )+…….+f (n )+……. Where f (n) decreases as n increases,

converges or diverges according as the integral ∫1

∞

f ( x )dx is finite or infinite.

p-series or Harmonic series test:

A positive term series ∑ un=∑ 1

np=1+ 1

2p+ 1

3p+……+ 1

np+…….is

i) Convergent if p>1ii) Divergent if p≤1

Proof:

Let f ( x )= 1

x p

∫1

∞

f ( x )dx=∫1

∞1

x pdx=[ x−p+1

−p+1 ]1

∞

, For p≠1

¿ { ∞ ,if−p+1>01p−1

, if−p+1<0

¿ { ∞ ,if p<11p−1

, if p>1

When p=1, ∫1

∞

f ( x )dx=∫1

∞1xdx= [logx ]1

∞=∞

Thus ∑ 1

np converges if p>1 and diverges if p≤1.

Theorem:

Let ∑ un be a positive term series. If ∑ un is convergent then limn→∞

un=0.

Proof:

If ∑ un is convergent then limn→∞

Sn=k.

un=(u1+u2+……+un )−(u1+u2+……+un−1 )

¿ Sn−Sn−1

limn→∞

Sn−1=k

limn→∞

un= limn→∞

Sn− limn→∞

Sn−1

¿k−k=0

Note:

Converse need not be always true. i.e. Even if limn→∞

un=0, then ∑ unneed not be convergent.

Example 1: ∑ un=1+ 12+ 1

3+ 1

4+…… ....

∑ un=1n

is divergent by integral test. But limn→∞

un= limn→∞

1n=¿0¿

Hence limn→∞

un=0 is a necessary condition but not a sufficient condition for convergence of

∑ un.

Example 2

Test the series for convergence, ∑n=2

∞1

nlogn

Solution: Consider ∫2

∞1

nlogndn=[ log ( logn ) ]2

∞=∞

Therefore ∑ un is divergent by Integral test.

Example 2

Test the series for convergence, ∑ ne−n2

Solution: Let x2=t . Then 2 xdx=dt

∫1

∞

x e−x2

dx=∫1

∞e−t

2dt=[ e−t−2 ]

1

∞

=1

2e

Therefore ∑ un is convergent.

Comparison test:

1. Let ∑ un and ∑ vn be two positive term series. If

a. ∑ vn is convergent

b. un≤vn ,∀n

Then ∑ un is also convergent.

That is if a larger series converges then smaller also converge.

2. Let ∑ un and ∑ vn be two positive term series. If

c. ∑ vn is divergent

d. un≥vn ,∀n

Then ∑ un is also divergent.

That is if a smaller series diverges then larger also diverges.

Example 2


∞1logn

Solution:

Let un=1

logn and vn=

1n

But ∑ vn=∑ 1n

is a p-series with p=1.

Therefore ∑ vn is divergent.

By comparison test ∑ un is also divergent.

Example 2

Test the series for convergence, ∑ 1

2n+1

Solution:

Let un=1

2n+1 and vn=

1

2n

But ∑ vn=∑ 1

2n is a geometric series with r=

12<1.

Therefore ∑ vn is convergent.

By comparision test ∑ un is also convergent.

Another form of comparison test is

Limit test

Statement: If ∑ un and ∑ vn be two positive term series such that limn→∞

unvn

=k (≠0). Then

∑ un and ∑ vn behave alike.

That is if ∑ un converges then ∑ vn also converge.

If ∑ un diverges then ∑ vn also diverge.

Examples 3.

Test the series for convergence, 1

1.2.3+ 3

2.3.4+ 5

3.4 .5+……….

Solution :

un=2n−1

n (n+1)(n+2)

Choose vn=1

n2 then limn→∞

unvn

=2

But ∑ vn=∑ 1

n2 with p=2>1.

Therefore ∑ vn is convergent. By limit test ∑ un is also convergent.

Examples 4.


∞

(√n2+1−n )

Solution: un=(√n2+1−n ) (√n2+1+n )(√n2+1+n )

¿n2+1−n2

√n2+1+n

¿1

n (√1+n2+1)

Let ∑ vn=∑ 1n

(p=1 )

limn→∞

unvn

=12

But ∑ vn is divergent. By limit test ∑ un is also divergent.

Examples 5.

Test the series for convergence, ∑ 3√n3+1−n

Solution:

Let ∑ vn=∑ 1

n2 with p=2>1.

limn→∞

unvn

=13

But ∑ vn is convergent. By limit test ∑ un is also convergent.

Example 6.

Test the series for convergence, Solve √2−133−1

+ √3−143−1

+ √4−153−1

+…………

Solution:

Let ∑ vn=∑ 1

n52 with p=

52>1.

limn→∞

unvn

=1


Example 7

Test the series for convergence, ∑ 1

n3tan

1n

Solution: un=1

n3tan

1n

We know that limn→∞

tan1n

1n

=1

Let ∑ vn=∑ 1

n4 . Then limn→∞

unvn

=1


Example 8

Test the series for convergence, ∑ 1n− log( n+1

n )

Solution:un=1n− log(1+ 1

n )¿ 1n−[ 1n− 1

2n2+ 1

6n3−……… ]

¿ [ 1

2n2− 1

6n3+………]

Let ∑ vn=∑ 1

n2 . Then limn→∞

unvn

=12


Exercises

Test for convergence of the series

1. ∑n=0

∞2n3+14n5+1

2. 1+ 22

2!+ 32

3 !+ 42

4 !+¿…… …∞

3.1

1.2.3+ 3

2.3.4+ 5

3.4 .5+¿…….. …∞

4. ∑ √ 3n−1

2n+1

5. ∑ nn

(n+1 )n+1

6.1

12+ 1+2

12+22+ 1+2+3

12+22+32+¿…… …∞

INFINITE SERIES

D’Alembert’s Ratio Test: If ∑ un is a series of positive terms, and ¿ l(a finite value)

then the series is convergent if l<1 , is divergent if l>1 and the test fails if l=1.

If the test fails, one should apply comparison test or the Raabe’s test, as given below:

Raabe’s Test: If ∑ un is a series of positive terms, and

limn→∞

n( unun+1

−1)=l ( finite ) , then the series is convergent if l>1 , is divergent if l<1 and the

test fails if l=1.

Remark: Ratio test can be applied when (i) vn does not have the form 1

np

(ii) nth term has xn , x2n etc.

(iii) nth term has n ! , (n+1 ) ! ,(n!)2 ect.

(iv) the number of factors in numerator and denominator increase steadily, ex: (

13+ 1.2

3.5+ 1.2.3

3.5 .7+…¿

Example : Test for convergence the series

1 +

22

2! +

32

3! +

42

4 ! + ….

>> The given series is of the form

12

1! +

22

2! +

32

3! +

43

4 ! + … whose nth term is un =

n2

n ! .

Therefore un+1 =

(n+1 )2

n+1 !

un+1

un =

(n+1 )2

n+1 !

n !

n2 =

(n+1 )2

n2.

n !(n+1 )(n !) =

n+1

n2

Therefore limn→∞

un+1

un = limn→∞

( n+1

n2 ) =

limn→∞

( 1n+

1

n2) = 0 < 1

Therefore by ratio test, un is convergent.

Example : Discuss the nature of the series

x1. 2 +

x2

2. 3 +

x3

3. 4 + ….

>> un =

xn

n(n+1 )

Therefore un+1 =

xn+1

(n+1 )(n+1+1) =

xn+1

(n+1 )(n+2)

Now

un+1

un =

xn+1

(n+1 )(n+2) .

n(n+1 )xn =

nn+2 x


un+1

un = limn→∞

nn+2 x =

limn→∞

1(1+2/n ) x = x

Therefore by D’Alembert’s ratio test un is {convergent if x < 1¿ ¿¿¿

And the test fails if x = 1

But when x = 1, un =

1n

n(n+1 ) =

1n(n+1 ) =

1

n2+n

un is of order 1/n2 (p = 2 > 1) and hence un is convergent (when x = 1). Hence we

conclude that un is convergent x 1 and divergent if x > 1

Example : Find the nature of series 1 +

x2 +

x2

5 +

x3

10 + ….

>> Omitting the first term, the given series can be written in the form

x1

12+1 +

x2

22+1 +

x3

32+1 + … so that un =

xn

n2+1

Therefore un+1 =

xn+1

n2+2n+2 .

n2+1n2+2n+2 x =

limn→∞

n2 (1+1 /n2 )n2(1+2 /n+2/n2 ) .x

That is, limn→∞

un+1

un = x

Hence by ratio test un is {convergent if x < 1¿ ¿¿¿

and the test fails if x = 1.

But when x = 1, un =

1n

n2+1 =

1

n2+1 is of order

1

n2 (p = 2 > 1)

Therefore un is convergent if x 1 and divergent if x > 1.

Example: Find the nature of the series

12√1 +

x2

3√2 +

x 4

4 √3 + …

>> omitting the first term, the general term of the series is given by un =

x2 n

(n+2 )√n+1

Therefore un+1 =

x2 (n+1)(n+1+2 )√(n+1)+1 =

x2 n+2

(n+3 )√n+2

un+1

un =

x2 n+2

(n+3 )√n+2

(n+2 )√n+1

x2 n

=

n+2n+3 √ n+1

n+2x2

=

√(n+2 )(n+1)(n+3 ) x2

limn→∞

un+1

un = limn→∞

√n(1+2/n)n(1+1/n )n(1+3/n ) . x2 = x2

Hence by ratio test un is {convergent if x2< 1 ¿¿¿¿

and the fails if x2 = 1.

When x2 = 1, un =

(1 )n

(n+2 )√n+1 =

1(n+2 )√n+1

un is of order 1/n3/2 (p = 3/2 > 1) and hence un is convergent.

Therefore un is convergent if x2 1 and divergent if x2 > 1.

Example : Discus the convergence of the series

x +

x3

2. 3 +

32. 4 +

x5

5 +

3 .52. 4 .6 .

x7

7 + … (x > 0)

>> We shall write the given series in the form

x +

12 .

x3

3 +

1 .32. 4 .

x5

5 +

1 .3 . 52. 4 .6 .

x7

7 + ….

Now, omitting the first term we have

un =

1. 3 .5 . ..(2n−1)2 . 4 . 6 .. .2n .

x2 n+1

2n+1

un+1 =

1. 3 .5 . ..[ 2(n+1 )−1 ]2. 4 .6 . .. 2(n+1 ) .

x2 (n+1)+1

2(n+1)+1

That is, un+1 =

1 .3 . 5. . .(2n+1)2. 4 .6 . .. .(2n+1) .

x2 n+3

2n+3

That is, un+1 =

1. 3 .5 . ..(2n−1)(2n+1)2 .4 .6 . . ..(2n )(2n+2) .

x2 n+3

2n+3

Therefore

un+1

un =

1. 3 .5 . ..(2n−1)(2n+1)2 .4 .6 . . ..(2n )(2n+2) .

x2 n+3

2n+3

2 . 4 . 6 .. .2n1. 3 .5 . ..(2n−1) .

2n+1

x2 n+1

That is,

un+1

un =

(2n+1)(2n+1 )x2

(2n+2)(2n+3 )


un+1

un = limn→∞

n(2+1/n )n (2+1/n) x2

n(2+2/n )n(2+3 /n) = x2

Hence by ratio test, un is {convergent if x2< 1 ¿¿¿¿

And the test fails if x2 = 1

When x2 = 1,

un+1

un =

(2n+1 )(2n+1 )(2n+2)(2n+3 ) and we shall apply Raabe’s test.

limn→∞ n

( unun+1

−1) =

limn→∞ n

[ (2n+2)(2n+3 )(2n+1)2

−1]

= limn→∞ n

[ (4n2+10n+6 )−(4 n2+4n+1 )(2n+1)2 ]

= limn→∞ n

( 6n+5

(2n+1 )2) =

limn→∞

n2 (6+5 /n)n2(2+1 /n)2

64 =

32 > 1

Therefore un is convergent (when x2 = 1) by Rabbe’s test.

Hence we conclude that, un is convergent if x2 1 and divergent if x2 > 1.

Example : Examine the convergence of

1 +

25 x +

69 x2 +

1417 x3 + … +

2n+1−22n+1+1 xn + ….

>> un =

2n+1−22n+1+1 xn.

Therefore un+1 =

2n+2−22n+2+1 xn+1

un+1

un =

2n+2−22n+2+1 xn+1

2n+1+12n+1−2 .

1

xn

un+1

un =

2n+2(1−2 /2n+2 )2n+2(1+1/2n+2 ) .x.

2n+1(1+1/2n+1 )2n+1(1−2 /2n+1 )

=

(1−1 /2n+1 )(1+1/2n+2 ) .x .

(1+1/2n+1 )(1−1 /2n)


un+1

un =

(1−0 )(1+0 ) . x .

(1+0 )(1−0 ) = x.

Therefore by ratio test un is {convergent if x < 1¿ ¿¿¿

and the test fails if x = 1.

When x = 1, un =

2n+1−22n+1+1

Therefore limn→∞ un =

limn→∞

2n+1(1−1 /2n)2n+1(1+1/2n+1 ) = 1

Since limn→∞ un = 1 0, un is divergent (when x = 1)

Hence un is convergent if x < 1 and divergent if x 1.

Example : test for convergence of the infinite series

1 +

2!

22 +

3!

33 +

4 !

44 + …

>> the first term of the given series can be written as 1!/11 so that we have,

un =

n !

nn and un+1 =

(n+1) !(n+1 )n+1

=

(n+1 )(n !)(n+1)n+1

=

n !

(n+1 )n

Therefore

un+1

un =

n !

(n+1 )n .

nn

n ! =

nn

(n+1 )n =

nn

nn (1+1 /n)n

limn→∞

un+1

un = limn→∞

1

(1+1/n )n =

1e < 1

Hence by ratio test un is convergent.

Cauchy’s Root Test: If ∑ un is a series of positive terms, and

limn→∞

(u¿¿n)1n=l(finite)¿,

then, the series converges if l<1, diverges if l>1 and fails if l=1.

Remark: Root test is useful when the terms of the series are of the form un=[ f (n )]g (n ).

We can note : (i) limn→∞

n1n=1

(ii) limn→∞

(1+ 1n)

1n=e

(iii) ) limn→∞

(1+ xn)

1n=e x

Example : Test for convergence ∑n=1

∞ [1+ 1√n ]

−n3/2

>> un = [1+ 1

√n ]−n3/2

Therefore (un)1/n = {[1+ 1

√n ]−n3 /2

}1 /n

= [1+ 1

√n ]−n1/2

= [1+ 1

√n ]−√n

limn→∞ (un) 1/n =

limn→∞

[1+ 1√n ]

−√n

= limn→∞

1

(1+ 1√n )

√n

=

1e < 1.

Therefore as n , √n also

Therefore by Cauchy’s root test, un is convergent.


∞

(1−3n )

n2

>> un = (1−3

n )n2

Therefore (un)1/n = [(1−3

n )n2

]1/n

= (1−3

n )n

limn→∞ (un)1/n =

limn→∞

(1+−34 )

n

= e-3.


(1+ xn )

n

= ex

That is, limn→∞ (un)1/n =

1

e3 < 1, therefore e = 2.7

Hence by Cauchy’s root test, un is convergent.

Example : Find the nature of the series ∑n=1

∞ [1+ 1√n ]

−n3 /2

>> un = [1+ 1

√n ]−n3/2

Therefore (un)1/n = {[1+ 1

√n ]−n3 /2

}1 /n

= [1+ 1

√n ]−n1/2

= [1+ 1

√n ]−√n


limn→∞

[1+ 1√n ]

−√n

= limn→∞

1

(1+ 1√n )

√n

=

1e < 1, since as n , √n also

Therefore by Cauchy’s root rest, un is convergent.


∞

(1−3n )

n2

>> un = (1−3

n )n2

Therefore (un)1/n = [(1−3

n )n2

]1/n

= (1−3

n )n


limn→∞

(1+−3n )

n

= e-3, since limn→∞

(1+ xn )

n

= ex

That is, limn→∞ (un)1/n =

1

e3 < 1, since e = 2.7.

Hence by Cauchy’s root test, un is convergent.

ALTERNATING SERIES

A series in which the terms are alternatively positive or negative is called an alternating

series.

i.e.,

LEBINITZ’S SERIES

An alternating series converges if

(i) each term is numerically less than its preceding term

(ii)

Note: If then the given series is oscillatory.

Q Test the convergence of

16 -

113 +

120 -

127 + …

Solution: Here un =

17n−1

then un+1 =

17(n=1)−1 =

17n+6

therefore, un – un+1 =

17n−1 -

17n+6

=

(7n+6 )−(7n−1 )(7n−1)(7n+6 ) =

7(7n−1)(7n+6 ) > 0

That is, un – un+1 > 0, un > un+1

Also, limn→∞ un =

limn→∞

17n−1 =

limn→∞

1n

1(7−1/n) = 0

Therefore by Leibnitz test the given alternating series is convergent .

Q Find the nature of the series

(1− 1log 2 )

- (1− 1

log 3 ) +

(1− 1log 4 )

- (1− 1

log 5 ) + …

Solution: Here un = 1 -

1log (n+1) then un+1 = 1 -

1log (n+2)

Therefore, un – un+1 =

1log (n+2) -

1log (n+1)

=

log (n+1)−log (n+2 )log (n+2) log (n+1) < 0.

Since (n + 1) < (n + 2)

un - un+1 < 0 un < un+1

further limn→∞ un =

limn→∞ 1 -

[ 1log(n+1 ) ]

= 1 – 0 = 1 0.

Both the conditions of the Leibnitz test are not satisfied. So, we conclude that the series

oscillates between - and + .

Problems:

Test the convergence of the following series

ABSOLUTELY & CONDITIONALLY CONVERGENT SERIES

An alternating series is said to be absolutely convergent if the positive

series is convergent.

An alternating series is said to be conditionally convergent if

(i) is divergent

(ii) is convergent

Theorem: An absolutely convergent series is convergent. The converse need not be true.

Proof: Let be an absolutely convergent series then is

convergent.

We know,

By comparison test, is convergent.

Q. Show that each of the following series also converges absolutely

(i) an2; (ii)

an2

1+an2

; (iii)

an1+an

Solution: (i) Since an converges, we have an 0 as n . Hence for some positive integer

N, |an| < 1 for all n N. This gives an2 |an| for all n N. As |an| is convergent it follows

an2 converges.

(as an2 is a positive termed series, convergence and absolute convergence are identical).

(ii) As 1 + an2 1 for all n, we get

an2

1+an2

an2

the convergence of an2 implies the convergence of

an2

1+an2

.

(iii) |an

1+an| =

|an||1+an| <

|an|1−|an| .

As |an| converges, |an| 0 as n . Hence for some positive integer N, we have |an| < ½

for all n N.

This gives |an

1+an| < 2|an| for all n N.

Now, by comparison test, |an

1+an| converges.

That is,

an1+an converges absolutely.

Q. Test the convergence

Solution: Here

then

i.e.,

Thus by Lebinitz rule, is convergent.

Also, . Take

Then

Since is divergent, therefore is also divergent.

Thus the given series is conditionally convergent.

POWER SERIES

A series of the form where the ’s are

independent of x, is called a power series in x. Such a series may converge for some or all

values of x.

INTERVAL OF CONVERGENCE

In the power series (i) we have

Therefore,

If , then by ratio test, the series (i) converges when and diverges for

other values.

Thus the power series (i) has an interval within which it converges and diverges

for values of x outside the interval. Such interval is called the interval of convergence of the

power series.

Q. Find the interval of convergence of the series .

Solution: Here and

Therefore,

By Ratio test the given series converges for and diverges for .

When x=1 the series reduces to , which is an alternating series and is

convergent.

When x=-1 the series becomes , which is divergent (by comparison with

p-series when p=1)

Hence the interval of convergence is .

Q. Show that the series ∑

1

∞

(−1 )n−1

xn

√2n+1 is absolutely convergent for | x | < 1,

conditionally convergent for x = 1 and divergent for x = -1.

Solution. Here un = (-1)n-1

xn

√2n+1

Therefore un+1 =

(−1)n xn+1

√2n+3

limn→∞

|uN+1

un| =

limn→∞

|(−1)n xn+1√2n+1

√2n+3(−1)n−1 xn|

= limn→∞

|(−1 )√ 2n+12n+3

x|

= limn→∞

|(−1 )√ n(2+1/n )n(2+3/n )

x| = | x |

Therefore by generalized D’ Alembert’s test the series is absolutely convergent if

| x | < 1, not convergent if | x | > 1 and the test fails if | x | = 1.

Now for |x | = 1, x can be +1 or – 1.

If x = 1 the given series becomes

1

√3 -

1

√5 +

1

√7 -

1

√9 + …

Here un =

1

√2n+1 , un+1 =

1

√2n+3

But 2n + 1< 2n + 3 un > un+1

Also limn→∞ un =

limn→∞

1

√2n+3 = 0

Therefore by Leibnitz test the series is convergence when x = 1.

But the absolute series

1

√3 +

1

√5 +

1

√7 + … whose general term is un =

1

√2n+1 and is of

order

1

√n =

1

n1/2 and hence un is divergent

Since the alternating series is convergent and the absolute series is divergent when x = 1, the

series is conditionally convergent when x = 1.

If x = -1, the series becomes

−1

√3 -

1

√5 -

1

√7 - ….

= - ( 1

√3+ 1

√5+ 1

√7+.. .)

where the series of positive terms is divergent as shown already.

Therefore the given series is divergent when x = -1.

Thus we have established all the results.

Problems:

1. Test the conditional convergence of

2. Prove that is absolutely convergent

3. For what values of x the following series are convergent

4. Test the nature of convergence

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infinite series.docx

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