Convergence / divergence examination $\sum_{k=1}^{\infty} \frac{\sin^{k}{\left(a \right)}}{k}$

$\sum_{k=1}^{\infty} \frac{\sin^{k}{\left(a \right)}}{k}$ Examination of a series depending on a real parameter a.

Can I use the squeeze theorem for series with the $\sum_{k=1}^{\infty} \frac{1}{k}$ series to judge this series and conclude that it is divergent? The exception would then be for a = 0, when the series is convergent. Or is the example solved completely differently and does it need some additional knowledge?

(I consider the absolute value of the sine, which is at most equal to 1)

(For the $\sum_{k=1}^{\infty} \frac{1}{k}$series, I assume divergence )

Solution 1:

• If $\sin(a)=0$, then the series obviously converges.

• If $\sin(a)=1$, then the series rewrites $\displaystyle{\sum \frac{1}{k}}$ which diverges.

• If $\sin(a)=-1$, then the series rewrites $\displaystyle{\sum \frac{(-1)^k}{k}}$ which converges by Alternating Test.

• If $|\sin(a)| < 1$ and $\sin(a) \neq 0$, then one has $$\left|\frac{\sin^{k+1}(a)/(k+1)}{\sin^k(a)/k}\right| = \left|\sin(a)\right| \times \frac{k+1}{k} \longrightarrow \left|\sin(a) \right|$$

Because $|\sin(a)| < 1$, then by the ratio test, the series converges.

In conclusion, the series converges iff $\sin(a) \neq 1$, i.e. $$\boxed{\text{the series converges iff } a \neq \dfrac{\pi}{2} \text{ mod } 2\pi}$$