Unit-3 SEQUENCES AND SERIES

3.1 Convergence of sequences and series.

Sequences:

Suppose that is a sequence. We say that if for every there is an N>0 so that whenever n>N,<. If we say that the sequence converges, otherwise it diverges.

Example1: Determine whether converges or diverges .If it converges, compute the limit.

Solution: we consider,

= 1-0 = 1

Thus the sequence converges to 1.

Example 2: Determine whether converges or diverges.If it converges, compute the limit.

Solution:

Now we consider,

Therefore by using L’Hospital’s rule. The sequence converges to 0.

Example 3: Suppose that and k is some constant

Solution:

= L+M

= L – M

,if M not equal to zero.

Series: A series is said to Converge to the sum s if the sequence of partial

Sums ( = 1 converges to s, i.e., =s.

*If the sum of the series is finite then the series is convergent.

Ie., 1 - .........

Example 1: Determine whether the following series converges or diverges

Solution: Consider the given series ie.,

= = - -

- ) =

Since , therefore the given series is convergent.

Example 2: Determine whether the following series is divergent or convergent

Solution: Given, s0 =1

s1 = 1-1=0

s2 = 1-1+1=1

s3 = 1-1+1-1=0

Hence the series diverges since doesn’t exist

P-series test:

p =.........

Converges , if p>1

Diverges , if p<1

Example 1:

=

Here p = 3 so p>1 ,thus the given series converges.

=

Here p= ie., p<1,thus the given series diverges

Example 2:

= +

= -3. +5.

=-3. +5.

Therefore., here p=2 ,3 ie., p>1

Hence the given series converges

Taylor series:

The taylor’s series can be represented as the following

(x-a)n

Example 1:

Find the taylor series for the following:

= <1

(x/10)<1 and (x/10) > -1

Therefore radius of convergence is (-10,10)

ROC =10

Example 2:

f(n)5 =

Here the ROC is 4

Exponential Series:

Example 1: test for convergence

Solution: given f(x) =

=

Thus, converges so by integral test also converges.

Example2: Solve for convergence.

Convergence for Logarithmic:

Example 1: f’(x) = ln() , >0 x

Solution:

Example2: Solve for convergence of the following

Solution:

Convergence for trigonometric functions:

Example:1: solve for convergence

Given

Note:

For small x values ,thus for large n’s we have,

Thus,

For large n’s

Thus,

Which clearly converges

Example2: Solve for convergence sin2x

Solution: we have sin2x =2sinx cosx…..(1)

Now, we find the convergence for the given trigonometric function.

Let 2x=u , then x =u/2 now we substitute these values in equation (1)

If 0<u< wwe can rewrite this as

But u<, then , and therefore

Let u=1. Since sin 1<1 , we find by using (*)

Let u = ½ . By using (*) again , and (1) we find that

Let u = 1/4 . By using(*) and (2) , we find that

Continue .in general we have

Thus

For 0 <x< , the since finction is an increasing function . It follows that

Example 1:

Using complex form, find the Fourier series of the function

f(x) = sinnx =

Solution:

We calculate the coefficients

=

=

Hence the Fourier series of the function in complex form is

We can transform the series and write it in the real form by renaming as

n=2k-1,n=

=

Example 2:

Using complex form find the Fourier series of the function f(x) = x2, defined on te interval [-1,1]

Solution:

Here the half-period is L=1.Therefore, the co-efficient c0 is,

For n

Integrating by parts twice, we obtain

=

=

= .

= .

Statement: If a function has a Fourier Series given by

Then Bessel’s inequality becomes an equality known as Parseval’s theorem .From (1)

Proof: We have

By squaring on both sides the above equation we get the following

Now by Integrating the above equation we get the following

So ,

For a generalized Fourier Series of a complete orthogonal system , an analogous relationship holds.

For a complex Fourier series,

Example : consider ,

Solution: The Fourier expansion is,

By Parseval’s formulae

is Reiman Zeta function defined by: