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Does the trigonometric series converge?

Examine if the series

\displaystyle \sum_{n=1}^{\infty} \sin \left[ \pi \left( 2+\sqrt{3} \right)^n \right]

converges.

Solution

The key lies in the fact that the number

\alpha = \left ( 2 + \sqrt{3} \right )^n +\left ( 2-\sqrt{3} \right )^n

is an integer. Let  x \in \mathbb{R} and let d(x) be the distance from x to the nearest integer. That is

d(x) = \left\{\begin{matrix} x & , & x \in \left [ 0, \frac{1}{2} \right ] \\\\ 1-x& , & x \in \left [ \frac{1}{2}, 1 \right ] \end{matrix}\right.

It is immediate that d can be expanded periodically with period T=1. Since \alpha is an integer we can get that

\displaystyle d\left [ \left ( 2 + \sqrt{3} \right )^n \right ] = \left ( 2-\sqrt{3} \right )^n = \frac{1}{\left ( 2+\sqrt{3} \right )^n}

Thus:

    \begin{align*} \left |\sum_{n=1}^{\infty} \sin \left [ \pi \left ( 2+\sqrt{3} \right )^n \right ] \right | &\leq \sum_{n=1}^{\infty} \left | \sin \left [ \pi \left ( 2+\sqrt{3} \right )^n \right ] \right | \\ &=\sum_{n=1}^{\infty} \sin \left [ \pi \; d\left ( 2+\sqrt{3} \right )^n \right ] \\ &\leq \pi \sum_{n=1}^{\infty} \frac{1}{\left ( 2+\sqrt{3} \right )^n} \end{align*}

and the conclusion follows.

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5 Comments

  1. There appears to be more in this series than meets the eyes. The convergence of this series is not exceptional! It is known that for almost all \alpha > 1 (i.e except a set of Lebesgue measure 0),
    \{ \alpha^n \}, the fractional part of \alpha^n is an equidistributed sequence. A consequence of this is for
    almost all \alpha > 1, the sequence \sin(\pi \alpha^n) does not converge to 0 and hence the series \sum\limits_{n=1}^\infty \sin(\pi \alpha^n) diverges.

    There are known exceptions to this. In particular, it is known that \{ \alpha^n \} is not equidistributed mod 1 if \alpha is a PV number, i.e. an algebraic integer \alpha > 1 and all other roots of its minimal polynomials lie strictly inside the unit circle.

    Hence the sum \sum\limits_{n=1}^\infty \sin(\pi\alpha^n) converges whenever \alpha is a PV number. Since 2 + \sqrt{3} is a PV number, the corresponding series do converge.

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