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In Exercises 52–57, do each of the following:

(a) Show that the given alternating series converges.

(b) Compute S10 and use Theorem 7.38 to find an interval containing the sum L of the series.

(c) Find the smallest value of n such that Theorem 7.38 guarantees that Sn is within 106 of L.

k=0(1)k(2k+1)!

Short Answer

Expert verified

(a). Using the test for convergence for alternating series we can see that the given alternating series converges.

(b). The value of S10 is 113!+15!17!+19!111!+113!115!+117!119!+121!0.841470984

and the value of L is L(0.84147098480789650665254,0.8414709848078965066526).

Step by step solution

01

Step 1. Given Information

Given, an alternating series

k=0(1)k(2k+1)!

02

Step 2. Using the alternating series test

Alternating series test for convergence is stated as if {ak} be a strictly decreasing sequence of positive numbers such that limkak=0 . Then the alternating series k=1(1)kak and k=1(1)k+1ak both converges.

03

Step 3. Applying the alternating series test to the given series 

The series k=01(2k+1)! is clearly a nonnegative strictly decreasing series. Also,

limk1(2k+1)!=0. Hence, by alternating series test this series converges and the series k=0(1)k(2k+1)! also converges.

04

Step 4. Given information

We are given with Theorem 7.8 to approximate the partial sum S10

05

Step 5. Calculating S10

If we add the first 10 terms of the series to obtain the 10th partial sum, S10, we obtain

S10=113!+15!17!+19!111!+113!115!+117!119!+121!0.841470984

This is an approximation for the sum L of the series. By Theorem 7.38, the error in using S10 is at most 123!3.86817×1023. In addition, Theorem 7.38 tells us that S10 is less than the sum of the series, L, because LS10has the same sign as the term 123!3.86817×1023 . Also,

LS10<a113.86817×1023. We have, L(S10,S10+a11). Therefore, L(0.84147098480789650665254,0.8414709848078965066526)

06

Step 6. Determining n

We have to find n such that

Sn<106

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