Sequence and Series

Mathematics (M101) Sequence and Series B.Tech 1st Year 1st Semester Sequence Definition: An Infinite sequence is a function f(n) whose domain is an infinite subset of whole numbers. We usually write a sequence {an} . where f(n) = an The following are the example of sequences {(.5)n} {n/(n+1)}

{en/n!} and {(-1)n}

You'll notice that sequences are discontinuous everywhere but never the less play an important role in Mathematics as we will see when we consider Infinite Series. 1. an = {(.5)n} 2. an = n/ (n+1) 3. an = {en/n!} 4. an= {(-1)n} As you could tell from the examples the 1st and 3d converge to 0, the second converges to 1 and the 4th diverges. But do we mean by convergence exactly? Definition:For any ε>0 there exists a number M such that L- ε < an < L + ε. whenever n> M Or as is usually stated given any ε>0 there exists a number M such that | an - L | < ε whenever n > M.Then the sequence an is said to converge to L. How do we compute limits? There are 2 very important theorems 1. Given {an} if there is a continuous differentiable function f(x) such that f(n) = an then lim an = lim f(x) therefore we can use our results on differentiable functions and L'Hopital's Rule. Examples 1 and 2 fit into this category. But what about the 3d? the Gamma Function

notwithstanding we need the following 2. If {an} is increasing and bounded above or decreasing and bounded below it converges. The first 3 examples make this fairly obvious.

an

Theorem: Suppose { } is a sequence which is increasing and bounded above, then it converges. an

Proof: Let L be the Least Upper Bound of { } i.e. it is an upper bound and there are none smaller. Let ε be given. Let M be the first number such that would be an Upper Bound. For all n>M

an

Further for all n

> an

am

since

L - ε . Such a number exists otherwise L -ε

is an increasing sequence.

because if L is an upper bound L+ ε is an upper bound.

Therefore we have shown given any ε > 0 There is a number M such that an

an

L-ε < < L+ ε whenever n > M i.e { } converges --In fact it converges to its Least Upper Bound.

Alternating Sequences: Another important theorem Suppose bn = (-1)n an where an >0. Then if lim an = 0 then lim bn = 0 If lim an ≠ 0 then { bn} diverges. Even if lim an = L The alternating sequence (-1)n an diverges as the subsequence of even terms converges to L and the subsequence of odd terms converges to - L therefore the sequence diverges. The next 2 examples show the divergent case and the 3d shows convergence of an alternating sequence.

1. (-1)n (n/(n+1)) 2. (-1)n * n 3. (-1)n (.5)n Let’s start off with some terminology and definitions. Given any sequence we have the following. 1. We call the sequence increasing if 2. We call the sequence decreasing if 3. If is an increasing sequence or monotonic.

for every n. for every n.

is a decreasing sequence we call it

4. If there exists a number m such that for every n we say the sequence is bounded below. The number m is sometimes called a lower bound for the sequence. 5. If there exists a number M such that for every n we say the sequence is bounded above. The number M is sometimes called an upper bound for the sequence. 6. If the sequence is both bounded below and bounded above we call the sequence bounded. 7. An infinite sequence is a function whose domain is the set of all positive integers. 8. A finite sequence is a function whose domain is {1, 2, 3, … , n}, the first n positive integers.

Examples: Determine if the following sequences are monotonic and/or bounded.

(a) { −n (b)

2

}

( −1)

n+1 ∞

(c)

 2   2    n  n = 5

Solutions: 1. This sequence is a decreasing sequence (and hence monotonic) because, for every n. Also, since the sequence terms will be either zero or negative this sequence is bounded above. We can use any positive number or zero as the bound, M, however, it’s standard to choose the smallest possible bound if we can and it’s a nice number. So, we’ll choose since,

This sequence is not bounded below however since we can always get below any potential bound by taking n large enough. Therefore, while the sequence is bounded above it is not bounded. As a side note we can also note that this sequence diverges (to specific).

if we want to be

2. The sequence terms in this sequence alternate between 1 and -1 and so the sequence is neither an increasing sequence or a decreasing sequence. Since the sequence is neither an increasing nor decreasing sequence it is not a monotonic sequence. The sequence is bounded however since it is bounded above by 1 and bounded below by -1. Again, we can note that this sequence is also divergent.

3. This sequence is a decreasing sequence (and hence monotonic) since,

The terms in this sequence are all positive and so it is bounded below by zero. Also, since the sequence is a decreasing sequence the first sequence term will be the largest and so we can see that the sequence will also be bounded above by sequence is bounded.

. Therefore, this

We can also take a quick limit and note that this sequence converges and its limit is zero. Convergence: an

A sequence { }of real (or complex) numbers is said to converge to a real (or complex) number c if for every > 0 there is an integer N > 0 such that if n > N then | an - c | < an The number c is called the alimit of the sequence { }and we sometimes write an c. n If a sequence { }does not converge, then we say that it diverges.

• •

Consider the sequence {1/n}. It converges to zero. The sequence {(-1)n}does not converge.

•

The sequence

Series|

converges to zero.

Definitions: Let { an } n =1 be a sequence. Define a new sequence { sn } n =1 , called the sequence ∞

∞

of partial sums for the sequence { an } n =1 , by the following equations: ∞

s1 = a1 , sn = sn −1 + an for n ≥ 2 . ∞

This new sequence of partial sums is called the series of an s, and it is denoted by ∑ an . n =1

∞

For each n ≥ 1 , sn is called the nth partial sum for the series ∑ an . The number an is called n =1

th

th

the n term of the series (as well as the n term of the sequence).  For n ≥ 2 , sn = sn −1 + an . This means s1 = a1 s2 = s1 + a2 = a1 + a2 s3 = s2 + a3 = a1 + a2 + a3 s4 = s3 + a4 = a1 + a2 + a3 + a4 M sn = sn −1 + an = a1 + a2 + a3 + L + an ∞

So we often write ∑ an = a1 + a2 + a3 + L + an + L n =1

 Sometimes the index for the sequence { an } n =1 does not begin with the number 1. In such cases the corresponding series is also indexed accordingly. For example, if ∞

∞

the sequence is { an } n =0 , then the series is ∑ an = a0 + a1 + a2 + L + an + L ; while if ∞

n =0 ∞

the sequence is { an } n =3 , then the series is ∑ an = a3 + a4 + a5 + L + an + L . ∞

n =3

Examples: Below are some examples that illustrate the difference between a sequence and the corresponding series and the sequence of partial sums. 1) Consider the sequence { ar

}

n ∞ n =0

, where a ≠ 0 and r ≠ 1 are constants. (If

r = 1 , then the sequence is the constant sequence of Example 1. If a = 0 , then the sequence is also a constant sequence of Example 1.)

Observe that { ar ∞

∑ ar

n

}

n ∞ n =0

= { a, ar , ar 2 , K , ar n , K } . The corresponding series is

= a + ar + ar 2 + L + ar n + L .

n =1

Calculating the sequence of partial sums: s0 = a0 = a s1 = s0 + a1 = a0 + a1 = a + ar

s2 = s1 + a2 = a0 + a1 + a2 = a + ar + ar 2 s3 = s2 + a3 = a0 + a1 + a2 + a3 = a + ar + ar 2 + ar 3 M sn = sn −1 + an = a0 + a1 + a2 + L + an = a + ar + ar 2 + L + ar n = a ( 1 − r n +1 ) ( 1 − r ) (This last equation is verified in class!) So the sequence of partial sums is

{ a ( 1− r ) n +1

( 1 − r ) } n=0 = { a, a ( 1 + r ) , a ( 1 + r + r2 ) ,K , a ( 1 − rn+1 ) ( 1 − r ) ,K } .

The sequence { ar

∞

}

∞

is called the Geometric Sequence and the series ∑ ar n =0

n ∞

n

n =0

is called the Geometric Series, and will be used often in this course. The term a is called the first term (for obvious reasons) and the number r is called the common ration since the ratio of consecutive terms is r . ∞

 1  2) Consider the sequence   .  n ( n + 1) n =1 ∞

 1  Observe that   = 1 2,1 6,1 12, K ,1  n ( n + 1)  , K . The  n ( n + 1) n =1

{

}

∞

1 1 1 1 1 = + + +L + + L . For any 2 6 12 n ( n + 1) n =1 n ( n + 1) integer k ≥ 1 , the following equality holds by partial fractions decomposition 1 1 1 = − (remember the technique of integration?) . We apply this k ( k + 1) k k + 1 equality to compute the partial sums as follows: 1 1 1 1 s1 = a1 = = − = 1− 1⋅ 2 1 2 2 1 1 1  1 1 1   1 1  s2 = s1 + a2 =  −  + =  −  +  −  = 1− 3 1 2  2 ⋅ 3  1 2   2 3  1  1 1  1  1 1  s3 = s2 + a3 = 1 −  + = 1 −  +  −  = 1 − 4  3 3⋅ 4  3 3 4  1  1 1  1 1 1 s4 = s3 + a4 = 1 −  + = 1 −  +  −  = 1 − 5  4 4⋅5  4 4 5 M 1 1  1  1  1 1 sn = sn −1 + an = 1 −  + = 1 −  +  − = 1−  n +1  n  n ( n + 1)  n   n n + 1  corresponding series is ∑

∞

1   So the sequence of partial sums is 1 −  . Notice that the equality from  n + 1 n =1 the partial fractions decomposition allows for the simplification of the partial

sums. Because of the fact that the intermediate terms add out, this series is called a telescoping series. ∞

Definitions: Let { an } n =1 be a sequence and let ∑ an denote the corresponding series. The ∞

n =1

∞

series ∑ an is said to converge provided the sequence of partial sums { sn } n =1 converges. ∞

n =1

∞

The series ∑ an is said to diverge provided the sequence of partial sums { sn } n =1 diverges. ∞

n =1 ∞

sn = S for some number S. This number S is The series ∑ an converges if and only if lim n →∞ n =1

∞

∞

n =1

n =1

called the sum of the series ∑ an and one writes ∑ an = S . Examples: The examples from above are reconsidered in the context of convergence/divergence. 1) Consider the Geometric Series ∞

∑ ar n =0

n

= a + ar + ar 2 + L + ar n + L ; a ≠ 0 , r ≠ 1 .

From above, the sequence of partial sums is

{ a ( 1− r ) n +1

( 1 − r ) } n=0 = { a, a ( 1 + r ) , a ( 1 + r + r2 ) ,K , a ( 1 − rn+1 ) ( 1 − r ) ,K } . ∞

(

)

a a lim ( 1 − rn +1 ) = 1 − lim rn +1 . From n →∞ n →∞ n →∞ 1 − r n →∞ 1− r n +1 r exists if and only if r < 1 our discussions on convergent sequences, lim n →∞ n +1 So lim sn = lim a ( 1 − r ) ( 1 − r ) =

∞

r n +1 = 0 for r < 1 , the series ∑ ar n converges (recall a ≠ 0 and r ≠ 1 ). Since lim n →∞ n =1

∞

a n if and only if r < 1 . In this case ∑ ar = . It is important to note that the 1− r n =1 Geometric Series converges if and only if the common ratio is less than one in absolute value. In this case, the sum of the Geometric Series is the first term of the series divide by the quantity one minus the common ratio. 2) Consider the telescoping series ∞ 1 1 1 1 1 = + + +L + +L . ∑ 2 6 12 n ( n + 1) n =1 n ( n + 1)

∞

1   From above, the sequence of partial sums is 1 −  . So  n + 1 n =1 1  1  lim sn = lim 1 − = lim1 − lim = 1 − 0 = 1 and the sequence of partial  n →∞ n →∞  n + 1  n →∞ n→∞ n + 1 ∞ ∞ 1 1 =1 sums converges to one. So the series ∑ converges and ∑ n =1 n ( n + 1) n =1 n ( n + 1) .

Now the question is- What properties of the terms an determine whether the series ∑ an converges or diverges? This question is answered by stating a number of ‘tests’ that can be applied to the terms an . The first such ‘test’ is the following: an ≠ 0 , then the series ∑ an The Divergence Test: Consider the series ∑ an . If lim n →∞ diverges. There are several observations one must keep in mind when applying the Divergence Test. an = 0 , then the series ∑ an converges. A. The test does not say that if lim n →∞

an = 0 , then one does not have any information about the B. If lim n →∞ convergence/divergence of the series. The Harmonic Series is an example of a an = 0 but the series diverges. The telescoping series above is an series with lim n →∞ an = 0 but the series converges. example of a series with lim n →∞ C. The Divergence Test can only show a series diverges. It never shows a series converges.

Examples: Below are examples of how we can show various series converge or diverge using the information we have so far. ∞

1) Does the series ∑ an converge or diverge for a ≠ 0 ? n =1

∞

an = a lim n ≠ 0 , the series ∑ an diverges by the Divergence Test. Since lim n →∞ n →∞ n =1

∞

2 2) Does the series ∑ 2 n converge or diverge? n=2 3

n n ∞ 2 2 1 1 for all n, the series = 2 = 2 ∑ 2 n is a Geometric Series  2   2n 3 n=2 3 3  9 2 2 2 1 with first term a = 2( 2) = 4 = and common ratio −1 < r = < 1 . So the 3 81 9 3 2 ∞ ∞ 2 2 a 1 = 81 = series ∑ 2 n converges and ∑ 2 n = . 1 − r 1 − 1 36 n=2 3 n=2 3 9

Since

2n 2 − 1 2 converge or diverge? n = 0 2 + 3n − 4n ∞

3) Does the series ∑

−2

2n − 1 2n − 1 n = lim ⋅ −2 2 2 n →∞ 2 + 3n − 4n n →∞ 2 + 3n − 4n n 2

2

Since lim

1 n2 = 2 = − 1 ≠ 0 , = lim n →∞ 2 3 2 + − 4 −4 2 n n 2−

2n 2 − 1 the series ∑ 2 diverges by the Divergence Test. n = 0 2 + 3n − 4n ∞

∞

1− n n 4) Does the series ∑ ( −1) 3 2 converge or diverge? n

n =1

Since ( −1) 31− n 2n = 3 ( −1) n

n

n

n ( 3−1 ) 2n = 3  − 23  for all n, the series 1

∞

n ( −1) 31−n 2n is a Geometric Series with first term a = 3  − 2  = −2 and ∑ n =1  3 ∞ n 1− n n 2 common ratio −1 < r = − < 1 . So the series ∑ ( −1) 3 2 converges and 3 n =1 ∞ a −2 6 n = = ( −1) 31−n 2n = ∑ 1− r  2 5 n =1 1−  −   3

Some important infinite series: Geometric Series: ∞

The series

∑x

n

is known as geometric series with common ratio x.

n =1

A Geometric series is convergent if -1