Chapter 7: Q. 24 (page 657)

Check the convergence $\sum_{k = 0}^{\infty} \frac{5^{k} + k}{k! + 3}$

Short Answer

Expert verified

Converges

Step by step solution

Step 1. Given

The given series $\sum_{k = 0}^{\infty} \frac{5^{k} + k}{k! + 3}$ .

Step 2. Ratio test

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

Step 3. Checking the convergence

$Now, calculate the value of \frac{a_{k + 1}}{a_{k}} Consider the general term a s = \frac{5^{k + 1}}{k! + 3} . . . (1 T h e r e f o r e, b y s u b s t i t u t i n g k = k + 1 \ln (1) a_{k + 1} = \frac{5^{(k + 1)} + (k + 1)}{(k + 1)! + 3} N o w, \frac{a_{k + 1}}{a_{k}} = \frac{5^{(k + 1)} + (k + 1) (k! + 3)}{(k + 1)! + 3 (5^{k + 1})} U \sin g f o r m u l a n! = n (n - 1)! \frac{a_{k + 1}}{a_{k}} = \frac{5^{(k + 1)} + (k + 1) (k! + 3)}{(k + 1)! + 3 (5^{k + 1})} T a k i n g l i m i t s \lim_{k \to \infty} \frac{a_{k + 1}}{a_{k}} = \lim_{k \to \infty} \frac{5^{(k + 1)} + (k + 1) (k! + 3)}{(k + 1)! + 3 (5^{k + 1})} L = 0 L < 1, The series converges$

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Check the convergence $\sum_{k = 0}^{\infty} \frac{5^{k} + k}{k! + 3}$

Short Answer

Step by step solution

Step 1. Given

Step 2. Ratio test

Step 3. Checking the convergence

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Check the convergence∑k=0∞5k+kk!+3

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

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Check the convergence $\sum_{k = 0}^{\infty} \frac{5^{k} + k}{k! + 3}$