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Chapter 7: Q. 20 (page 639)

Let $伪 and 尾$ be two distinct positive numbers less than $1$ . Explain why the ratio test cannot be used on the series $伪 + 尾 + 伪^{2} + 尾^{2} + 伪^{3} + 尾^{3} + 鈰�$ . Then show that the series converges and find its sum.

Short Answer

Expert verified

Hence proved.

Step by step solution

01

Step 1, Given information.

The given series is $伪 + 尾 + 伪^{2} + 尾^{2} + 伪^{3} + 尾^{3} + 鈰�$

02

Step 2. Conclusion.

We can rewrite the series as,

$(伪 + 伪^{2} + 伪^{3} 鈥�) + (尾 + 尾^{2} + 尾^{3} + 鈥�)$

localid="1649162284800" $Rewrite 伪 (1 + 伪 + 伪^{2} 鈥�) = 伪 {鈭�}_{k = 0}^{鈭�} \frac{1}{伪^{k}} And 尾 (1 + 尾 + 尾^{2} + 鈥�) = 尾 {鈭�}_{k = 0}^{鈭�} \frac{1}{尾^{k}} T h e r e f o r e, (伪 + 伪^{2} + 伪^{3} 鈥�) + (尾 + 尾^{2} + 尾^{3} + 鈥�) = 伪 {鈭�}_{k = 0}^{鈭�} \frac{1}{伪^{k}} + 尾 {鈭�}_{k = 0}^{鈭�} \frac{1}{尾^{k}} H e n c e, t h e s e r i e s c o n v e r g e s . N o w, t h e s u m i s, 伪 {鈭�}_{k = 0}^{鈭�} \frac{1}{伪^{k}} + 尾 {鈭�}_{k = 0}^{鈭�} \frac{1}{尾^{k}} is 伪 (\frac{1}{1 - 伪}) + 尾 (\frac{1}{1 - 尾}) 伪 (\frac{1}{1 - 伪}) + 尾 (\frac{1}{1 - 尾}) = (\frac{伪}{1 - 伪}) + (\frac{尾}{1 - 尾})$

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