 rocedwrp

2021-03-06

Find the interval of convergence of the power series.
$\sum _{n=0}^{\mathrm{\infty }}\frac{{x}^{5n}}{n!}$ avortarF

Consider the given power series:
$\sum {a}_{n}=\sum _{n=1}^{\mathrm{\infty }}\frac{{x}^{5n}}{n!}$
Here the objective is to find the interval of x for which the given power series is convergent.
According to the ratio test
$L=\underset{n\to \mathrm{\infty }}{lim}|\frac{{a}_{n+1}}{{a}_{n}}|$
If $L<1$ then the series converges absolutely
If $L>1$ then the series is divergent
Here ${a}_{n}=\frac{{x}^{5n}}{n!}$ Replace $n\to n+1$
${a}_{n+1}=\frac{{x}^{5n+5}}{\left(n+1\right)!}$
Use ratio test for convergence
$L=\underset{n\to \mathrm{\infty }}{lim}|\frac{{a}_{n+1}}{{a}_{n}}|$
Substitute ${a}_{n+1}=\frac{{x}^{5n+5}}{\left(n+1\right)!}$ and ${a}_{n}=\frac{{x}^{5n}}{n!}$
$L=\underset{n\to \mathrm{\infty }}{lim}|\frac{\frac{{x}^{5n+5}}{\left(n+1\right)!}}{\frac{{x}^{5n}}{n!}}|$
$L=\underset{n\to \mathrm{\infty }}{lim}|\frac{n!}{{x}^{5n}}\frac{{x}^{5n+5}}{\left(n+1\right)!}|$
$L=\underset{n\to \mathrm{\infty }}{lim}|\frac{n!}{{x}^{5n}}\frac{{x}^{5n}×{x}^{5}}{\left(n+1\right)\left(n\right)!}|$
$L=\underset{n\to \mathrm{\infty }}{lim}|\frac{{x}^{5}}{\left(n+1\right)}|$
$L=|{x}^{5}|\underset{n\to \mathrm{\infty }}{lim}\frac{1}{\left(n+1\right)}$
$L=|{x}^{5}|×0$
$L=0<1$
Here the limit is less than 1, and independent of the value of x.
Hence the given power series is convergent for all $x\in \left(-\mathrm{\infty },\mathrm{\infty }\right)$ Jeffrey Jordon