91Ó°ÊÓ

Use the ratio test to determine the radius of convergence (with ):$a_n=\frac{(n+2)!}{n!}$

$\begin{array}{c}\underset{n→\mathrm{∞}}{\lim }\left|\frac{a_n}{a_{n+1}}\right|=\underset{n→\mathrm{∞}}{\lim }\left|\frac{\frac{(n−2)!}{n!}}{\frac{(n+3)!}{(n+1)!}}\right|\\ =\underset{n→\mathrm{∞}}{\lim }\left|\frac{(n+2)!}{(n+3)!}â‹\dots \frac{(n+1)!}{n!}\right|\\ =\underset{n→\mathrm{∞}}{\lim }\left|\frac{(n+2)!}{(n+3)(n+2)!}×\frac{(n+1)n!}{n!}\right|\\ =\underset{n→\mathrm{∞}}{\lim }\left|\frac{(n+1)}{n+3}\right|\\ =\underset{n→\mathrm{∞}}{\lim }\left|\frac{(1+(1/n\left)\right)}{1+(3/n)}\right|\\ =1\end{array}$

The radius of convergence is 1, therefore convergent set for the given power series is.$|x+2|<1$

To completely identify the convergence set, we have to check whether the boundary points -1 and -3 are included in the set or not.

Checking at$x=−1$, by substituting the value ofx by -1 .

$\begin{array}{c}\underset{n=0}{\overset{\mathrm{∞}}{∑}}\frac{(n+2)!}{n!}{\left(x+2\right)}^n=\underset{n=0}{\overset{\mathrm{∞}}{∑}}\frac{(n+2)!}{n!}{\left(-1+2\right)}^n\\ =\underset{n=0}{\overset{\mathrm{∞}}{∑}}\frac{(n+2)!}{n!}{\left(1\right)}^n\\ =\underset{n=0}{\overset{\mathrm{∞}}{∑}}\frac{(n+2)(n+1)n!}{n!}\\ =\underset{n=0}{\overset{\mathrm{∞}}{∑}}(n+2)(n+1)\\ =\mathrm{∞}\end{array}$

The above series is divergent, thus the point -1 is excluded from the convergent set

Similarly, checking at $x=−3$by substituting the value of x by -3.

$\begin{array}{c}\underset{n=0}{\overset{\mathrm{∞}}{∑}}\frac{(n+2)!}{n!}{\left(x+2\right)}^n=\underset{n=0}{\overset{\mathrm{∞}}{∑}}\frac{(n+2)!}{n!}{\left(-3+2\right)}^n\\ =\underset{n=0}{\overset{\mathrm{∞}}{∑}}\frac{(n+2)!}{n!}\\ =\underset{n=0}{\overset{\mathrm{∞}}{∑}}\frac{(n+2)(n+1)n!}{n!}\\ =\underset{n=0}{\overset{\mathrm{∞}}{∑}}{(−1)}^n(n+2)(n+1)\end{array}$

The above series is divergent, thus the point -3 is not included in the convergent set. The convergent set for the given power series is.$x∈(−3,−1)$