"Does the series∑n=1∞sin⁡(2πn2+a2sin⁡n+(−1)n)converge?" was answered by students like you

sweetrainyday8s2

Answered question

2022-02-24

Does the series

\sum_{n = 1}^{\infty} \sin (2 π \sqrt{n^{2} + a^{2} \sin n + {(- 1)}^{n}})

converge?

Donald Erickson

Beginner2022-02-25Added 8 answers

This series is convergent.
As n tends to

+ \infty

, we may write

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

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

= \sin (2 π n (1 + \frac{a^{2} \sin n}{2 n^{2}} + \frac{{(- 1)}^{n}}{2 n^{2}} + O (\frac{1}{n^{4}})))

= \sin (2 π n + \frac{π a^{2} \sin n}{n} + \frac{π {(- 1)}^{n}}{n} + O (\frac{1}{n^{3}}))

= \sin (\frac{π a^{2} \sin n}{n} + \frac{π {(- 1)}^{n}}{n} + O (\frac{1}{n^{3}}))

= \frac{π a^{2} \sin n}{n} + \frac{π {(- 1)}^{n}}{n} + O (\frac{1}{n^{3}})

Now recall that

\sum_{n = 1}^{+ \infty} \frac{\sin n}{n}

is convergent, moreover

\sum_{n = 1}^{+ \infty} \frac{\sin n}{n} = I \sum_{n = 1}^{+ \infty} \frac{e^{\in}}{n} = I (- \log (1 - e^{i})) = \frac{π - 1}{2}

Then it is clear that your initial series

\sum_{n = 1}^{+ \infty} u_{n}

is convergent, being the sum of convergent series.

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