Sequences And Series Of Real Numbers -3 - IIT JAM MCQ

If S_n denote the sum of n terms of the series  then S_n will be > Explanation:

A: Sn > n^2
- The given comparison tells us that the sum of the first n terms of the series is greater than n^2.
- This could be true for certain series where the sum of terms grows faster than n^2, such as series with exponential growth.
- However, this comparison alone does not provide enough information to determine the behavior of the series for all values of n.

B: Sn > n
- This comparison indicates that the sum of the first n terms of the series is greater than n itself.
- Similar to the first comparison, this could be true for certain series with increasing terms.
- Again, this comparison alone does not give a definitive answer about the convergence or divergence of the series.

C: Sn = ∞
- This comparison suggests that the sum of the terms of the series diverges to infinity.
- If the sum of the series approaches infinity as n increases, it indicates that the series does not converge.
- This comparison implies that the series may be divergent and does not have a finite sum.

D: Sn = ∞
- This comparison reiterates the previous point that the sum of the series diverges to infinity.
- It reinforces the idea that the series does not have a finite sum and diverges as n increases.

Conclusion:
- Based on the comparisons provided, it is most likely that the series described does not converge and has an infinite sum. This is indicated by the comparisons stating that the sum of the terms exceeds n^2 and n, and ultimately diverges to infinity.

Let the given expansion be f(n)

f(n) = (1 + a₁)(1 + a₂)... (1 + a_n)

Also given, a₁ = a₂ = a₃ = ... = a_n = 1

Consider for n = 2

f(2) = (1 + a₁)(1 + a₂)

f(2) = (1 + 1)(1 + 1) = 2²

Consider for n = 5

f(5) = (1 + a₁)(1 + a₂)(1 + a₃)(1 + a₄)(1 + a₅)

f(5) = (1 + 1)(1 + 1)(1 + 1)(1 + 1)(1 + 1) = 2⁵

Similary for n times, it is given as

f(n) = (1 + 1)(1 + 1) ...... (1 + 1) = 2ⁿ

(1 + a₁)(1 + a₂)... (1 + a_n) = 2ⁿ

Since, we know that ifbe a sequence such that  and 


Hence, 
and 

If m > n,

Since 2r – 1 ≤ r !

Therefore, it follows that <s_n> is a cauchy sequence. Hence <s_n> converges
 

The set of all cluster points of a sequence is sometimes called the limit set. contains all but finitely many elements of the sequence). That is why we do not use the term limit point of a sequence as a synonym for accumulation point of the sequence.

A sequence that is not a Cauchy sequence is called a divergent sequence. A Cauchy sequence is a sequence whose elements get closer and closer together as the sequence progresses. 

just divide numertor and denominator by n² 

• A Cauchy sequence is one where the terms become arbitrarily close to each other as the sequence progresses.
• In the real numbers (R), every Cauchy sequence converges to a limit. This means A is true.
• A Cauchy sequence is also bounded, as the terms cannot diverge infinitely far apart. Hence, B is also true.
• Therefore, both statements A and B are true, making the correct answer option 1.

The correct option is 2.

Since Sa_n is convergent, its sequence of terms must converge to 0. This is because if the terms did not converge to zero, then the sum would not converge.

On the other hand, Sb_n is divergent, which means that its sequence of terms does not converge to 0. If it did, then the series would converge by the n^th term test.

Therefore, option 2 is the correct answer.

For any positive number ε there is a natural number n, such that
​

For C, suppose lim a_2n = p, lim a_2n+1 = q, lim a_3n = r.

Considering a_6n, p = r.

Considering a_6n+3 = a_2(3n+1)+1, q = r.

Therefore p=q=r.

Since a_2n and a_2n+1 include all integers, limits exists.

So C is true.