Let San be a convergent series of positive terms and let Sbn be a divergent series of positive terms. Then,a)the sequence is convergent and is not convergentb)the sequence converges to 0c)the sequence does not converge to 0d)the sequence diverges to∞Correct answer is option 'B'. Can you explain this answer?

Question

Let San be a convergent series of positive terms and let Sbn be a divergent series of positive terms. Then,a)the sequence < an > is convergent and < bn> is not convergentb)the sequence < an > converges to 0c)the sequence < bn > does not converge to 0d)the sequence < bn > diverges to∞Correct answer is option 'B'. Can you explain this answer?

Answer

Answer is abc

Answer

Take an example of =1/n^2 and =1/n

then you can omit options a,c.d

Answer

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.

Answer

The position to sequence and to all keep here and to all
keep so we're and to all among to his so we all
b is correct answer

Answer

The sequence {(San + Sbn)} is divergent.

Proof:
Since San is a convergent series, we know that the sequence of partial sums of San, {Sn}, is convergent. Let L be the limit of {Sn}, i.e., lim(n→∞) Sn = L.

Now, consider the sequence {(San + Sbn)}. The partial sum of this sequence, denoted by {Tn}, can be written as Tn = San + Sbn.

Since San is convergent, we have lim(n→∞) San = 0 (by the definition of convergence). Similarly, since Sbn is divergent, we have lim(n→∞) Sbn = ∞ (by the definition of divergence).

Now, let's consider the limit of {Tn}:
lim(n→∞) Tn = lim(n→∞) (San + Sbn)

Since lim(n→∞) San = 0 and lim(n→∞) Sbn = ∞, we can rewrite the above expression as:
lim(n→∞) Tn = 0 + ∞ = ∞

Therefore, the sequence {(San + Sbn)} is divergent.

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