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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 < ∞, then either both the series converge or both the series diverge. Here, the uniformity in the nature of convergence or divergence of both the series will change, if we consider the case that L = 0, thus making the theorem to lose its validity. We shall consider the following example to prove this claim. Let an be the series n=1∞1n2 . we all know that this series is a convergent series. Our bn should be such that the ratio of their limits is 0. Hence, let bn be n=1∞1n. This is a harmonic series and a divergent one also. The limit L = limn→∞1n21n = limn→∞nn21 = limn→∞1n = 0. Here, an converges and bn diverges, making the theorem to lose its base.
The one side of the theorem which is always true, irrespective of the nature of the limit is the antecedent part of the theorem albeit with a small change. It says that if an and bn are two series with an, bn ≥ 0 for all n, then if lim⁡supn→∞anbn = L, with 0 ≤ L < ∞, then both the series converges. The concept of limit superior is necessary here, as if we take the limit alone, then the theorem may not be necessarily true, as the limit of a convergent sequence exists imply limit superior = limit inferior = the original limit.
Let an = (1- (-1)n) / n2 and bn = 1/n2 for all n > 1. Thus, limn→∞anbn = limn→∞(1- (-1)n) doesn’t exist. Now consider, lim⁡supn→∞anbn = lim⁡supn→∞ an / lim⁡supn→∞ bn= lim⁡supn&a...
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