Question

$\sum _{n=1}^{\mathrm{\infty }} (\sin x{)}^n(-1{)}^n{2}^n$

Short answer

The interval of convergence is

$\left(-\frac{\pi }{6},\frac{\pi }{6}\right)$

Step-by-step solution

 Step 1: Given Information

The power series is $\sum _{n=1}^{\mathrm{\infty }} (\sin x{)}^n(-1{)}^n{2}^n$

Definition of the interval of convergence

The Interval of convergence is the interval in which the power series is convergent.

Find the interval.

The power series is $\sum _{n=1}^{\mathrm{\infty }} (\sin x{)}^n(-1{)}^n{2}^n$

Let $y=\sin x$

The power series becomes $\sum _{n=1}^{\mathrm{\infty }} y^n(-1{)}^n{2}^n$

$\text{Let}{\rho }_n=\left|\frac{a_{n+1}}{a_n}\right|\text{,}$

Substitute the value of the power series in the formula above, the equation becomes as follows,

$\begin{array}{r}{\rho }_n=\left|\frac{a_{n+1}}{a_n}\right|\\ =\left|\frac{y^{n+1}{2}^{n+1}}{y^n{2}^n}\right|\\ =\left|2y\right|\end{array}$

Apply limits in the above equation,

$\begin{array}{r}\rho =\underset{n\to \mathrm{\infty }}{lim} \left|2y\right|\\ =\left|2y\right|\end{array}$

The power series is convergent for $\rho <1$

Hence the given power series is convergent for$\left|2y\right|<1\text{and divergent for}\left|2y\right|>1\text{.}$.

The series is convergent when $\left|2y\right|<1$

$\begin{array}{r}\left|2y\right|<1\\ -\frac{1}{2}<y<\frac{1}{2}\end{array}$

Find the limits of x,

$\begin{array}{r}-\frac{1}{2}<\sin x<\frac{1}{2}\\ {\sin }^{-1}-\frac{1}{2}<x<{\sin }^{-1}\frac{1}{2}\\ n\pi -\frac{\pi }{6}<x<n\pi +\frac{\pi }{6}\end{array}$

Now check the ends points,

When $x=-\frac{\pi }{6}\text{series is}\sum _{n=1}^{\mathrm{\infty }} 1^n$, which is a divergent series.

When $x=\frac{\pi }{6}\text{series is}\sum _{n=1}^{\mathrm{\infty }} (-1{)}^n$, which is a divergent series.

Hence the interval of convergence is$\left(-\frac{\pi }{6},\frac{\pi }{6}\right)$