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


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.

real analysis II
answers for the questions
* Given that the limit of the sequence {an} is 0. Let us consider the series n=0∞an. 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=0∞an2. To find the nature of this series, let us take the famous ratio test for convergence. For this case, we have limn→∞|an+1an |. Now, limn→∞|an+1an | = limn→∞|an+12an2 | > 1 (As we don’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=0∞|an |. Consider limn→∞|an|. Given that limn→∞an is 0. The terms of √|an| < an. Which means limn→∞|an| is also 0. By the theorem, if limn→∞an is 0 for a series n=1∞an , then the series is convergent, we have n=0∞|an to be convergent.
* Consider the series n=1∞ann. Let the series bn = n=1∞1n. Now, the terms ann ≤ 1n. But the series n=1∞1n is divergent by the theorem, n=1∞1np is divergent if p ≤1, we have the series n=1∞ann 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 ≥ 0 for all n, then if limn→∞anbn = L, with 0 &l...
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