Unit-3

Sequence & Series

Question-1: If , then the limit of will be,

Ans 1) = = = ½

Hence the limit of the sequence is 1/2.

Question-2: check whether the series is convergent or divergent. Find its value in case of convergent.

Ans 2) As we know that,

Sn =

Therefore,

Sn =

Now find out the limit of the sequence,

= ∞

Here the value of the limit is infinity, so that the series is divergent as sequence diverges.

Question-3: check whether the series is convergent or divergent.

Ans 3). The general formula can be written as,

We get on applying limits,

) = 3/4

This is the convergent series and its value is 3 / 4

Question-4: Test the convergence of the following series-

Ans 4) Here we have the series,

Now,

Now comapare

We can see that the limit is finite and not zero.

Here and converges or diverges together since ,

is the form of here p = 1,

So that,

is divergent then is also divergent.

Question-5: Test the convergence of the series whose n’th term is given below-

n’th term =

Ans 5)

We have and

By D’Alembert ratio test,

So that by D’Alembert ratio test , the series will be convergent.

Question-6: If the series converges, then find the value of x.

Ans 6) Here

Then,

By D’Almbert’s ratio test the series is convergent for |x|<1 and divergent if |x|>1.

So at x = 1

The series becomes-

At x = -1

This is an alternately convergent series.

This is also convergent series, p = 2

Here, the interval of convergence is

Question-7: Express the polynomial in powers of (x-2).

Ans 7). Here we have,

f(x) =

differentiating the function w.r.t.x-

f’(x) =

f’’(x) = 12x + 14

f’’’(x) = 12

f’’’’(x)=0

now using Taylor’s theorem-

+ ……. (1)

Here we have, a = 2,

Put x = 2 in the derivatives of f(x), we get-

f(2) =

f’(2) =

f’’(2) = 12(2)+14 = 38

f’’’(2) = 12 and f’’’’(2) = 0

now put a = 2 and substitute the above values in equation(1), we get-

Question-8: Find the Taylor’s expansion of about (1 , 1) up to second degree term.

Ans 8). We have,

At (1 , 1)

Now by using Taylor’s theorem-

……

Suppose 1 + h = x then h = x – 1

1 + k = y then k = y - 1

……

=

……..

Question-9: Verify

Ans 9). As we know that

And

Hence,

and

That means

Question-10: Find half range cosine series of in the interval and hence deduce that

Ans 10)

Here

;

Hence it’s half range cosine series is,

… (1)

Where

Hence equation (1) becomes,

… (2)

Put x = 0, we get

Hence the result

Put we get,

i.e.

Question-11: Prove that for 0 < x <

1.

2.

Sol. 1. Half range series,

=

= 0 when n is odd

So that-

Now by Parseval’s formula-

So that-