Sequences And Series Of Real Numbers -2 - IIT JAM MCQ

So, we can see that from S₁ and S₂ that is divergent and S₂ is convergent.

The nth term of the given series is

{S_n} is increasing sequence and unbounded from above.
∴ It is divergent sequence.
Hence, the series is divergent series.

If a_1^, a₂……… are in AP and b₁, b₂………. are in GP then a₂b₂, a₂b₂,……… are in AGP.

Here, given series is

or 0 as n is odd or even implies {S_n} - { 1 , 0 , 1 , 0 , . . . }
It is an oscillating sequence

Hint:
Let first term is a and the common difference is d of the AP
Now, T_m = 1/n
⇒ a + (m-1)d = 1/n ………… 1
and T_n = 1/m
⇒ a + (n-1)d = 1/m ………. 2
From equation 2 – 1, we get
(m-1)d – (n-1)d = 1/n – 1/m
⇒ (m-n)d = (m-n)/mn
⇒ d = 1/mn
From equation 1, we get
a + (m-1)/mn = 1/n
⇒ a = 1/n – (m-1)/mn
⇒ a = {m – (m-1)}/mn
⇒ a = {m – m + 1)}/mn
⇒ a = 1/mn
Now, T_mn = 1/mn + (mn-1)/mn
⇒ T_mn = 1/mn + 1 – 1/mn
⇒ T_mn = 1

Here

The series x  is convergent.
If -1 <x ≤ 1.

Here, it is given that

For ratio test
∑a_n is convergent if

here it is given that T3 = 4.

⇒ ar² = 4
Now product of first five terms = a.ar.ar².ar³.ar⁴

= a⁵r¹⁰
= (ar²)^5
= 4⁵

and the radius of convergence

∴ 

By ratio test



Hence, the series diverges for | x | > 1.

We have 
implies 
Hence, series is convergent.

We have 
a_n = (-1)ⁿ 

implies {a_n} is a bounded sequence.

implies { b_n } is converges to zero.
Since, {a_n} is bounded sequence and {b_n} converges to zero.
Hence, {a_nb_n} converges to zero.

and the radius of convergence

This problem is a very basic one, this problem can easily be solved by step by step solution. The steps are:

Step 1 : First we will ignore the summation part. We will factorize the denominator, because we are going step by step so our aim is to simplify the given problem first.

Step 2: After factorizing the the denominator we will reach to a position where we have to use partial fraction to go forward.

Step 3: In this step we will take care of the (−1)�� part, like how it will affect the series.

Step 4: After taking care of the (−1)�� we will now expand the summation (breaking it into infinite sum).

Step 5 : So after 4 steps we are halfway done now just the last simplification is left we will use the value

ln2=1−1/2+1/3−1/4+…

 is convergent, therefore 
 is also convergent.