solutions for quiz xxix

In sum 2n lim n 1 2n 2 4 1 2 the radius

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Unformatted text preview: = ≤ = 2 3 2n 2 3 2n 2 3 2n 2 3 2n 2 3 = 2 4 · · 2n 3 3 2n 3 3 + 2 · 32 · 2n 3 4 + 1 32 2n 2 2n 3 5 + 2 + + 1 52 + 1 22 −1− 3 6 3 6 2n 2 + ··· + ··· 1 62 + 2 π · 32 · 6 + ··· 3 5 + 1 42 + 2n 2 6 + 3 4 + 2n 2 5 + −→ − + ··· 0 as n goes to ∞. In sum, ζ 2n lim n− ∞ → ζ −1 2n + 2 = 4. −1 ∞ (2) The radius of convergence for ζ 2n n=1 Work : Consider the series with variable u: ∞ ζ 2n n=1 12 −1 un . −1 x2 n is R = 2. Set an = ζ 2n − 1. =ζ an =⇒ Then an+1 lim n− ∞ → 2n − 1, = ζ 2n + 2 − 1. ζ an an+1 = lim n− ∞ → 2n ζ 2n + 2 −1 , −1 which equals 4, by virtue of the result of (1). Hence the radius of convergence for the power series ∞ ( ∗) ζ 2n n=1 un −1 equals 4. This implies that the radius of convergence for the original series ∞ ζ 2n −1 √ x2 n 4 = 2. n=1 equals R= Indeed, the original series with variable x i...
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