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Question Number 197323 by Erico last updated on 13/Sep/23

If f (x) = \frac{\sin (x)}{x} and S_{n} (α) = \underset{k = 1}{\sum^{n}} [f (k π + \frac{π}{α}) + f (k π - \frac{π}{α})] (α > 1)

Prove that \lim_{n \to + \infty} S_{n} (α) = 1 - f (\frac{π}{α})

Answered by witcher3 last updated on 14/Sep/23

S_{n} (a) = \underset{1}{\sum^{n}} (f (k π + \frac{π}{a}) - f (k π - \frac{π}{a}))

T (a) \underset{k = 1}{\sum^{n}} \frac{{(- 1)}^{k} \sin (\frac{π}{a})}{k π + \frac{π}{a}}

- T (- a) + T (a) = S_{n} (a)

T (a) = \frac{\sin (\frac{π}{a})}{π} \underset{k = 1}{\sum^{n}} \frac{{(- 1)}^{k}}{k + \frac{1}{a}} \dots cv

Σ {(- 1)}^{k} ⩽ 2, . . \frac{1}{k + \frac{1}{a}} decrease cv \to 0

T (a) . . cv

\underset{k = 1}{\sum^{\infty}} \frac{{(- 1)}^{k}}{k + \frac{1}{a}} = \underset{k = 1}{\sum^{\infty}} \frac{1}{2 k + \frac{1}{a}} - \frac{1}{2 k - 1 + \frac{1}{a}}

= \frac{1}{2} \underset{k = 1}{\sum^{\infty}} \frac{1}{k + \frac{1}{2 a}} - \frac{1}{k + \frac{1}{2} (\frac{1}{a} - 1)}

= \frac{1}{2} (Ψ (\frac{1}{2} (\frac{1}{a} + 1)) - Ψ (\frac{1}{2 a} + 1))

S_{n} (a) = Γ (a) - Γ (- a)

= \frac{\sin (\frac{π}{a})}{2 π} (Ψ (\frac{1}{2} + \frac{1}{2 a}) - Ψ (\frac{1}{2} - \frac{1}{2 a}) - Ψ (1 + \frac{1}{2 a}) + Ψ (1 - \frac{1}{2 a}))

= \frac{\sin (\frac{π}{a})}{2 π} (Ψ (1 - (\frac{1}{2} - \frac{1}{2 a}) - Ψ (\frac{1}{2} - \frac{1}{2 a}) - 2 a - Ψ (\frac{1}{2 a}) + Ψ (1 - \frac{1}{2 a}))

= \frac{\sin (\frac{π}{a})}{2 π} (π \cot (π (\frac{1}{2} - \frac{1}{2 a}) - 2 a + π \cot (\frac{π}{2 a})

= \frac{\sin (\frac{π}{a})}{2} (tg (\frac{1}{2 a}) + \cot (\frac{1}{2 a})) - \frac{\sin (\frac{π}{a})}{\frac{π}{a}}

= \frac{\sin (\frac{π}{a})}{2} (\frac{\cos^{2} (\frac{1}{2 a}) + \sin^{2} (\frac{1}{2 a})}{\sin (\frac{π}{2 a}) \cos (\frac{π}{2 a})}) - f (\frac{π}{a})

= 1 - f (\frac{π}{a})

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