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Chapter 7: Q. 18 (page 639)

Explain why the series ${鈭�}_{k = 0}^{鈭�} \frac{1}{k!}$ converges. Which convergence tests could be used to prove this?

Short Answer

Expert verified

Hence proved.

Step by step solution

01

Step 1. Given Information.

The given series is ${鈭�}_{k = 0}^{鈭�} \frac{1}{k!}$ .

02

Step 2. Ratio Test.

Using the ratio test,

$\frac{a_{k + 1}}{a_{k}} = \frac{\frac{1}{(k + 1)!}}{\frac{1}{k!}} \frac{a_{k + 1}}{a_{k}} = \frac{k!}{(k + 1) k!} = \frac{1}{k + 1} \lim_{k 鈫� 鈭�} \frac{a_{k + 1}}{a_{k}} = \lim_{k 鈫� 鈭�} \frac{1}{(k + 1)} = 1 (\lim_{k 鈫� 鈭�} \frac{1}{(k + 1)}) = 0 S i n c e, L < 1,$

Therefore, the series converges.

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Most popular questions from this chapter

Determine whether the series ${鈭�}_{k = 2}^{鈭�} {(\frac{- 3}{5})}^{k}$ converges or diverges. Give the sum of the convergent series.

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Use the divergence test to analyze the given series. Each answer should either be the series diverges or the divergence test fails, along with the reason for your answer.

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Given that $a_{0} = - 3, a_{1} = 5, a_{2} = - 4, a_{3} = 2$ and ${鈭� a_{k}}_{k = 2}^{鈭�} = 7$ , find the value of ${鈭� a_{k}}_{k = 0}^{鈭�}$ .

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