study-the-convergence-n-o-sin-pi-4n-2-2-

Question Number 199707 by SANOGO last updated on 08/Nov/23

s t u d y t h e c o n v e r g e n c e

\underset{n ⩾ o}{\sum^{}} s i n (π \sqrt{4 n^{2} + 2}

Answered by witcher3 last updated on 09/Nov/23

\sqrt{{4 n}^{2} + 2} = 2 n \sqrt{1 + \frac{1}{{2 n}^{2}}}

\sin (2 π n \sqrt{1 + \frac{1}{{2 n}^{2}}}) = \sin (2 π n \sqrt{1 + \frac{1}{{2 n}^{2}}} - 2 π n)

= \sin (2 π n (\sqrt{1 + \frac{1}{{2 n}^{2}}} - 1)

\sqrt{1 + \frac{1}{{2 n}^{2}}} - 1 = \frac{1}{{2 n}^{2} (\sqrt{1 + \frac{1}{{2 n}^{2}}} + 1)} ⩽ \frac{1}{{4 n}^{2}}

\sin (0) ⩽ \sin (2 π n \sqrt{1 + \frac{1}{{2 n}^{2}}} - 2 π n) ⩽ \sin (\frac{π}{{2 n}^{2}}) ⩽ \sin (\frac{π}{2})

positiv serie

2 π n (\sqrt{1 + \frac{1}{{2 n}^{2}}}) - 2 π n

= 2 π n (1 + \frac{1}{{2 n}^{2}} - 1 + o (\frac{1}{n^{2}}) = \frac{π}{n} + o (\frac{1}{n})

\sin (\frac{π}{n} + o (\frac{1}{n})) \sim \frac{π}{n}; Σ \frac{π}{n} dv \Rightarrow Σ \sin (π \sqrt{{4 n}^{2} + 2}) dv

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