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-