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# Real Analysis II Assignment: Answers for the Questions (Math Problem Sample)

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this file is a continuation of the previous article. that is part 1 and this is part 2. as said above, the solutions are my own and hence, no sources.

source..Content:

real analysis II

answers for the questions

1

* Given that the limit of the sequence {an} is 0. Let us consider the series n=0âˆžan. To find the nature of this series, let us consider the sequence of the nth partial sum of the series say {sn}. Then sn = a0+a1+...+an. Then according to the theorem, if the sequence of the partial sum of the series converges to L, then the series itself will converge to L, we can say that the given series converges because, the sequence of the partial sums here are nothing but the terms of the convergent sequence say a0, a1, a2 and so on. The sequence itself is a convergent sequence. Then, their partial sums also converge, which implies that the series itself is convergent and converges to 0.

* Now consider the series, n=0âˆžan2. To find the nature of this series, let us take the famous ratio test for convergence. For this case, we have limnâ†’âˆž|an+1an |. Now, limnâ†’âˆž|an+1an | = limnâ†’âˆž|an+12an2 | > 1 (As we donâ€™t know the nature of the sequence {an2}). Here, only the sequence {an} is given as convergent. Since the terms an2 will be positive, their absolute value will obviously be positive and thus the ratio between two absolute values will normally be greater than 1. This means that the given series is divergent.

* consider the series n=0âˆž|an |. Consider limnâ†’âˆž|an|. Given that limnâ†’âˆžan is 0. The terms of âˆš|an| < an. Which means limnâ†’âˆž|an| is also 0. By the theorem, if limnâ†’âˆžan is 0 for a series n=1âˆžan , then the series is convergent, we have n=0âˆž|an to be convergent.

* Consider the series n=1âˆžann. Let the series bn = n=1âˆž1n. Now, the terms ann â‰¤ 1n. But the series n=1âˆž1n is divergent by the theorem, n=1âˆž1np is divergent if p â‰¤1, we have the series n=1âˆžann to be divergent.

2 first we shall state the Limit Comparison theorem. It says that if an and bn are two series with an, bn â‰¥ 0 for all n, then if limnâ†’âˆžanbn = L, with 0 &l...

answers for the questions

1

* Given that the limit of the sequence {an} is 0. Let us consider the series n=0âˆžan. To find the nature of this series, let us consider the sequence of the nth partial sum of the series say {sn}. Then sn = a0+a1+...+an. Then according to the theorem, if the sequence of the partial sum of the series converges to L, then the series itself will converge to L, we can say that the given series converges because, the sequence of the partial sums here are nothing but the terms of the convergent sequence say a0, a1, a2 and so on. The sequence itself is a convergent sequence. Then, their partial sums also converge, which implies that the series itself is convergent and converges to 0.

* Now consider the series, n=0âˆžan2. To find the nature of this series, let us take the famous ratio test for convergence. For this case, we have limnâ†’âˆž|an+1an |. Now, limnâ†’âˆž|an+1an | = limnâ†’âˆž|an+12an2 | > 1 (As we donâ€™t know the nature of the sequence {an2}). Here, only the sequence {an} is given as convergent. Since the terms an2 will be positive, their absolute value will obviously be positive and thus the ratio between two absolute values will normally be greater than 1. This means that the given series is divergent.

* consider the series n=0âˆž|an |. Consider limnâ†’âˆž|an|. Given that limnâ†’âˆžan is 0. The terms of âˆš|an| < an. Which means limnâ†’âˆž|an| is also 0. By the theorem, if limnâ†’âˆžan is 0 for a series n=1âˆžan , then the series is convergent, we have n=0âˆž|an to be convergent.

* Consider the series n=1âˆžann. Let the series bn = n=1âˆž1n. Now, the terms ann â‰¤ 1n. But the series n=1âˆž1n is divergent by the theorem, n=1âˆž1np is divergent if p â‰¤1, we have the series n=1âˆžann to be divergent.

2 first we shall state the Limit Comparison theorem. It says that if an and bn are two series with an, bn â‰¥ 0 for all n, then if limnâ†’âˆžanbn = L, with 0 &l...

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