Question

In Problems 1-10 find the interval and radius of convergence for the given power series.

$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\frac{{\mathbf{5}}^{\mathbf{n}}}{\mathbf{n}\mathbf{!}}{\mathit{x}}^{\mathbf{n}}$

Short answer

The radius of convergence is$R=\infty \text{interval:}(-\infty ,\infty )$

Step-by-step solution

Ratio test

The Ratio test for the power series,$\sum _{n=1}^{\infty }{a}_n{(x-a)}^n$is given as,

$\underset{\mathbf{n}\mathbf{\to }\mathbf{\infty }}{\mathbf{l}\mathbf{i}\mathbf{m}}\left|\frac{{\displaystyle a_{n+1}{(x-a)}^{n+1}}}{{\displaystyle a_n{(x-a)}^n}}\right|\mathbf{=}\mathit{L}$

If, $L<1$then the series converges absolutely.

If, $L>1$then the series diverges.

If, L=1 then the test is inconclusive.

Apply ratio test

Consider the series$\sum _{n=1}^{\infty }\frac{5^n{x}^n}{n!}$

The ratio test gives;

$\begin{array}{c}\underset{n\to \infty }{\lim }\left|\frac{\frac{5^{n+1}{(x-0)}^{n+1}}{(n+1)!}}{\frac{5^n{\left(x\right)}^n}{\left(n\right)!}}\right|=\underset{n\to \infty }{\lim }\left|\frac{5^{n+1}{x}^{n+1}}{(n+1)!}\cdot \frac{n!}{5^n{x}^n}\right|\\ =\underset{n\to \infty }{\lim }\left|\frac{5^n\cdot 5\cdot x^n\cdot x}{(n+1)n!}\cdot \frac{n!}{5^n{x}^n}\right|({\text{sincea}}^{m+n}=a^m.)\\ =\underset{n\to \infty }{\lim }\left|\frac{\frac{5}{(n\cdot 5\cdot x}\cdot x}{(n+1)\gg !}\cdot \frac{{\displaystyle \prod }^{\text{'}n}}{5^n\times }\right|\left(\text{since}\right(n+1)!=(n+1)\cdot n!)\\ =\underset{n\to \infty }{\lim }\left|\frac{1}{(n+1)}\cdot \frac{5x}{1}\right|\\ =0\end{array}$

For radius & interval of convergence

For convergence we need.$0<1$ However, this is always true x is.

Thus the radius of convergence and the interval are$R=\infty \text{interval:}(-\infty ,\infty )$