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Chapter 8: Q. 21 (page 692)

Use an appropriate Maclaurin series to find the values of the series in Exercises 17鈥�22.

鈭�k=0鈭�(-1)k蟺2k+1(2k+1)!

Short Answer

Expert verified

The required answer is鈭�k=0鈭�(-1)k蟺2k+1(2k+1)!=0

Step by step solution

01

Step 1. Given Information    

The given series is鈭�k=0鈭�(-1)k蟺2k+1(2k+1)!

02

Step 2. Explanation    

The maclaurin series for the function f(x)=sinxissinx=鈭�k=0鈭�(-1)kx2k+1(2k+1)!

So, the given series is the maclaurin series for sinxatx=蟺

Since, 鈭�k=0鈭�(-1)kx2k+1(2k+1)!=sinx

Thus,

鈭�k=0鈭�(-1)k蟺2k+1(2k+1)!=蝉颈苍蟺鈭�k=0鈭�(-1)k蟺2k+1(2k+1)!=0

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Most popular questions from this chapter

Find the interval of convergence for power series:鈭�k=0鈭�1k!xk

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