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(a) Show that if the sequence (Sn) converges to $, then the sequence {s; } converges to $? (b) Use (a) and the fact that for all x,y R,X> = 4Ir + y)? (= 9)?] to ...

Question

(a) Show that if the sequence (Sn) converges to $, then the sequence {s; } converges to $? (b) Use (a) and the fact that for all x,y R,X> = 4Ir + y)? (= 9)?] to give an alter- native proof of Theorem 4.2.2(ii) .

(a) Show that if the sequence (Sn) converges to $, then the sequence {s; } converges to $? (b) Use (a) and the fact that for all x,y R,X> = 4Ir + y)? (= 9)?] to give an alter- native proof of Theorem 4.2.2(ii) .



Answers

Let $\sum a_{n}$ be a series that converges to a number $L$ and whose terms are all positive. We show that any rearrangement $\sum c_{n}$ of this series also converges to $L$. (a) Show that $S_{n}<L$ for any partial sum $S_{n}$ of $\sum c_{n}$ (b) Show that for any $K<L,$ there is a partial sum $S_{n}$ of $\sum c_{n}$ with $S_{n}>K$ (c) Use parts (a) and (b) to show $\sum c_{n}$ converges to $L$

In discussion we need to show that sequins And upon and plus one converges to the number L. is equals 2. 1. Let's see how the soldiers question. Consider A. And it equals two and upon and plus one. In order to find the limit of sequins, we can right limit and turns to infinity A. n equals two limit and tends to infinity And upon and plus one we can write this limit and Kansas infinity. Wanna bone one plus one upon end. Now substitute and is equal to infinity. So we get one of 11 plus zero and this will be calls to one hands. We can say that. Yeah. The limit of the sequence A. n. equals two and upon and plus one is one. Yeah. Does the sequence kind of orange juice? Two. Number L. difficult to one. This is the final answer for this problem. I hope you understood the solution. Thank you.

In this question we need to show that chick wins into the power and plus one upon he took the power and convergence to number L. is the calls to one. Let's see how to solve this question. Consider A. N. If he calls to eat with the power and plus one upon into the power end. In order to find the limit of the sequence. We can right limit and tends to infinity A. And vehicles too limit and tends to infinity into the power and plus one upon U. To the power. And or we can write the right hand side as limit and tends to infinity one plus one upon U. To the power. And or this will be calls to limit And turns to Infinity one plus limit and tends to infinity one upon it with the power and no substitute. And is the cost to infinity. So we get 1-plus zero and this will be called to one hands. The limit of sequence limit of sequence A and F equals two E. To the power and plus one upon Italy power and is one. Therefore the sequence converges two L equals to one. So this is the final answer for this problem. I hope you understand resolution. Thank you


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