Q 55. - Calculus

Question

Explain why the series is not a power series in $x-x_0$.Then use the ratio test for absolute convergence to find the values of $x$ for which the given series converge  $\sum _{k=0}^{\infty }\frac{{\left(\sin x\right)}^k}{k!}$

Step-by-Step Solution

Answer

The value of $x$ for which the series $\sum _{k=0}^{\infty }\frac{{\left(\sin x\right)}^k}{k!}$converges when $x\in R$.

Step 1. Given information.

The given power series is $\sum _{k=0}^{\infty }\frac{{\left(\sin x\right)}^k}{k!}$

Step 2. Find the values of x for which the given series converge.

Since, the series consists of power of $Sin x$. So the series in not power series in $x-x_0$.

Now, 

$b_k=\frac{{\left(\sin x\right)}^k}{k!}$ and $b_{k+1}=\frac{{\left(\sin x\right)}^{k+1}}{\left(k+1\right)!}$

Thus,

$$\begin{gathered} \underset{k\to \infty }{\lim }\left|\frac{b_{k+1}}{b_k}\right|=\underset{k\to \infty }{lim}\left|\frac{{\displaystyle \frac{{\left(\sin x\right)}^{k+1}}{\left(k+1\right)!}}}{{\displaystyle \frac{{\left(\sin x\right)}^k}{k!}}}\right| \\ =\underset{k\to \infty }{lim}\left|\frac{\sin x}{\left(k+1\right)}\right| \end{gathered}$$

So, for $k\to \infty$the value of limit will always be zero.

Therefore, the value of $x$ for which the series $\sum _{k=0}^{\infty }\frac{{\left(\sin x\right)}^k}{k!}$ converges when $x\in R$.