Q. 29 Check the convergence聽鈭慿=0鈭�... [FREE SOLUTION]

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Chapter 7: Q. 29 (page 657)

Check the convergence ${鈭�}_{k = 0}^{鈭�} \frac{1}{2^{k} + \sin k}$

Short Answer

Expert verified

Converges

Step by step solution

Step 1. Given

The given series is ${鈭�}_{k = 0}^{鈭�} \frac{1}{2^{k} + \sin k}$ .

Step 2. Limit comparison test

$According to limit comparison test, let {鈭�}_{k = 1}^{鈭�} a_{k} a n d {鈭�}_{k = 1}^{鈭�} b_{k} be two series with positive terms 1 . If \lim_{k 鈫� 鈭�} \frac{a_{k}}{b_{k}} = L, where L is any positive real number, then either both series converges or both series diverges . 2 . \lim_{k 鈫� 鈭�} \frac{a_{k}}{b_{k}} = 0, {鈭�}_{k = 1}^{鈭�} b_{k} converges then {鈭�}_{k = 1}^{鈭�} a_{k} diverges 3 . \lim_{k 鈫� 鈭�} \frac{a_{k}}{b_{k}} = 鈭�, {鈭�}_{k = 1}^{鈭�} b_{k} diverges then {鈭�}_{k = 1}^{鈭�} a_{k} Converges$

Step 3. Checking the convergence

Consider the sequence $a_{k} = \frac{1}{2^{k} + \sin k} a n d b_{k} = \frac{1}{2^{k}} \lim_{k 鈫� 鈭�} \frac{a_{k}}{b_{k}} = \lim_{k 鈫� 鈭�} \frac{\frac{1}{2^{k} + \sin k}}{\frac{1}{2^{k}}} s olving this limit = \frac{1}{1 + 0} = 1 {鈭�}_{k = 1}^{鈭�} b_{k} = {鈭�}_{k = 1}^{鈭�} \frac{1}{2^{k}} The given serie s {鈭�}_{k = 1}^{鈭�} b_{k} = {鈭�}_{k = 1}^{鈭�} \frac{1}{2^{k}} is of the form {鈭�}_{k = 1}^{鈭�} b_{k} = {鈭�}_{k = 1}^{鈭�} r^{k}, where |r| < 1 . From limit comparison test the series converges .$

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Check the convergence ${鈭�}_{k = 0}^{鈭�} \frac{1}{2^{k} + \sin k}$

Short Answer

Step by step solution

Step 1. Given

Step 2. Limit comparison test

Step 3. Checking the convergence

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Check the convergence鈭�k=0鈭�12k+sink

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Step by step solution

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Check the convergence ${鈭�}_{k = 0}^{鈭�} \frac{1}{2^{k} + \sin k}$