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Determine the radius of convergence of ∑ (sin (an))zn

by RoRi

Determine the radius of convergence r of the power series:

    \[ \sum_{n=0}^{\infty} (\sin (an))z^n \qquad a> 0. \]

Test for convergence at the boundary points if r is finite.

If |z| \geq 1 then \lim_{n \to \infty} (\sin (an))z^n \neq 0 unless a = k \pi. Thus, the series does not converge for a \neq k \pi if |z| \geq 1.

If |z| < 1 then we apply Dirichlet’s test where

    \[ \sum_{n=1}^N | \sin (an) | \]

is bounded for any N (i.e., the partial sums are bounded) and the z^n terms are monotonically decreasing with \lim_{n \to \infty} z^n = 0. Hence, the series converges for |z| < 1, which implies r = 1 if a \neq k \pi.

If a = k \pi then \sin (an) = 0 for all n so the series converges for all z which implies r = +\infty.

One comment

  1. MathlessRick says:

    This | sine | series diverges though

    You could have said it’s going to be less or equal to |z|^n which converges if |z| 1 but this is no problem since, as I said above, | sin(ax) | diverges when |z| = 1, so… yeah

    Btw, the sine series diverges because the limit of the absolute value doesn’t go to zero (see theorem 10.6)

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