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:
