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N 1 1 ∞ ∼ , and n=1 is divergent, it follows from the criterion of equivalence that n2 + 1 n n b) is divergent. It also follows that neither a) nor c) can be absolutely convergent. Since a n → 0 for n → ∞, we must apply Leibniz’s criterion. Clearly, both series are alternating. If we put 1) Since ϕ(x) = x2 x , +1 er ϕ (x) = x2 +1−2x2 1 − x2 = <0 2 2 (x +1) (1+x2 )2 for x > 1, then ϕ(x) → 0 decreasingly for x → ∞, x > 1. Then it follows from Leibniz’s criterion that both a) and c) are (conditionally) convergent.

Each term of the Fourier series is continuous, while the sum function f ∗ (t) is not continuous. Hence, it follows that the Fourier series cannot be uniformly convergent in R. 3) When F ∈ K2π , then F is periodic of period 2π, hence π π F (t) dt = −π 0 F (t) dt + 0 π F (t + 2π) dt = 2π F (t) dt + −π 0 2π F (t) dt = π F (t) dt. 0 In particular, an = 1 π 2π t2 cos nt dt, og 0 bn = 2π 1 π t2 sin nt dt. 0 Thus a0 = 1 π 2π t2 dt = 0 8π 2 8π 3 = , 3π 3 and an = 2π 1 π t2 cos nt dt = 0 = 0+ 1 2 t sin t πn 2 2 [t cos nt]2π 0 − 2 πn πn2 2π 2 2π − 0 πn 2π cos nt dt = 0 t sin nt dt 0 2 4 · 2π = 2 2 πn n for n ≥ 1, and bn = 1 π = − 2π t2 sin nt dt = 0 1 −t2 cos nt πn 4π 2 2 2 + [t sin nt]2π 0 − πn πn2 πn2 2π 0 ∞ 4π 2 + 3 n=1 + 2 πn 2π t cos nt dt 0 sin nt dt = − The Fourier series is (NB.

4n2 − 1 (2n − 1)(2n + 1) 2 2n − 1 2 2n + 1 The corresponding segmental sequence is then N sN ∞ ∞ = 1 1 1 1 1 = − 2 2 n=1 2n − 1 2 n=1 2n + 1 4n − 1 n=1 = 1 1 1 − 2 n=1 2n − 1 2 N N +1 n=2 1 1 1 1 = − · 2n − 1 2 2 2N + 1 1 → for N → ∞, 2 and the series is convergent with the sum ∞ = lim sN = n=1 N →∞ 1 . com 53 Examples of Fourier series Sum function of Fourier series b) When we insert t = π into the Fourier series, we get f (π) = 1 = 2 4 − π π ∞ 4 (−1)n = 2−1 4n π n=1 ∞ (−1)n−1 1 + 2 n=1 4n2 − 1 .

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