Radius and Interval of Convergence for (x/sin(n))^n

[Denis99]

Find Radius and Interval of Convergence for $$\sum_{1}^{\infty}(\frac{x}{sinn})^{n}$$.
I don`t have any ideas how to do that :/

 

[HOI]

The standard method of determining the radius of convergence of power series is to use the "ratio test": \left|\left(\frac{x^{n+1}}{sin^{n+1}(n+1)}\right)\left(\frac{sin^n(n)}{x^n}\right)\right|= |x|\frac{sin^n(n)}{sin^{n+1}(n+1)}.

The radius of convergence is 1 over the limit of the fraction.

 

[Denis99]

The standard method of determining the radius of convergence of power series is to use the "ratio test": \left|\left(\frac{x^{n+1}}{sin^{n+1}(n+1)}\right)\left(\frac{sin^n(n)}{x^n}\right)\right|= |x|\frac{sin^n(n)}{sin^{n+1}(n+1)}.

The radius of convergence is 1 over the limit of the fraction.

Thank you for your answer :)
I was thinking about this, but is there a way to calculate limit of this fraction? Sinuses are problematic for me in this limit.

 

[Opalg]

Find Radius and Interval of Convergence for $$\sum_{1}^{\infty}(\frac{x}{sinn})^{n}$$.
I don't have any ideas how to do that :/

The ratio test won't work in this example. I think you will find that in this case the radius of convergence is zero. To see that, use the "$n$th term test": if the $n$th term of a series does not tend to zero then the series does not converge.

Informally, the argument goes like this. The value of $\sin n$ varies in a fairly random way in the interval $[-1,1]$, but from time to time it gets very close to zero. Now suppose that $x\ne0$. For infinitely many values of $n$, $\sin n$ will be small enough that $\left|\dfrac x{\sin n}\right| >1$. That shows that $\left(\dfrac x{\sin n}\right)^n \not\to0$, so the series $$\sum_{n=1}^\infty \left(\dfrac x{\sin n}\right)^n$$ does not converge.