Q. 28 Check the convergence聽鈭慿=0鈭�... [FREE SOLUTION]

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Chapter 7: Q. 28 (page 657)

Check the convergence ${鈭�}_{k = 0}^{鈭�} \frac{5^{k}}{(k + 1)!}$

Short Answer

Expert verified

Converges

Step by step solution

Step 1. Given

The given series is ${鈭�}_{k = 0}^{鈭�} \frac{5^{k}}{(k + 1)!}$ .

Step 2. Ratio test

$According to the RatioTest, {鈭� a_{k}}_{k = 1}^{鈭�} be the series with positive terms, if L = \lim_{k 鈫� 鈭�} \frac{a_{k + 1}}{a_{k}} If aL < 1 series converges . 2 . If L > 1 series diverges . 3 . If L = 1 the test is inconclusive .$

Step 3. Checking the convergence

$Now, calculate the value of \frac{a_{k + 1}}{a_{k}} Consider the general term a s = \frac{5^{k}}{(k + 1)! .} . . . (1 T h e r e f o r e, b y s u b s t i t u t i n g k = k + 1 \ln (1) a_{k + 1} = \frac{5^{(k + 1)}}{((k + 1) + 1)!} = \frac{5^{(k + 1)}}{((k + 2)!} N o w, \frac{a_{k + 1}}{a_{k}} = \frac{\frac{5^{k + 1}}{(k + 2)!}}{\frac{5^{k}}{(k + 1)!}} U \sin g f o r m u l a n! = n (n - 1)! \frac{a_{k + 1}}{a_{k}} = \frac{5}{k + 2} T a k i n g l i m i t s \lim_{k 鈫� 鈭�} \frac{a_{k + 1}}{a_{k}} = \lim_{k 鈫� 鈭�} (\frac{5}{k + 2}) L = 0 T h e s e r i e s c o n v e r g e s$

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91影视

Check the convergence ${鈭�}_{k = 0}^{鈭�} \frac{5^{k}}{(k + 1)!}$

Short Answer

Step by step solution

Step 1. Given

Step 2. Ratio test

Step 3. Checking the convergence

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91影视

Check the convergence鈭�k=0鈭�5k(k+1)!

Short Answer

Step by step solution

Step 1. Given

Step 2. Ratio test

Step 3. Checking the convergence

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Probability and Statistics

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Study anywhere. Anytime. Across all devices.

Check the convergence ${鈭�}_{k = 0}^{鈭�} \frac{5^{k}}{(k + 1)!}$