Find the radius of convergence of $$\sum_{n=0}^\infty\left(\frac{-1}{27}\right)^n(x-3)^{3n+2}$$
Attempt:
I set $d_{3n} = 0$, $d_{3n+1} = 0$, $d_{3n+2} = a_n$, $d_{3n+3} = 0$
Hence the series becomes $\sum_{n=0}^\infty d_n(x-3)^n$
Now, its given that $R=\limsup(b_n)^{1/n}$ in the book
here, $R=\limsup(d_n)^{1/n}=\limsup(a_n)^{3/(n-2)}$
That is, $R=\limsup\left(\frac{-1}{27}\right)^{3n/(n-2)}$
How will I proceed further? (How can I evaluate the lim sup)
Letting $y = x-3$,
$\begin{array}\\ \sum_{n=0}^\infty\left(\frac{-1}{27}\right)^n(x-3)^{3n+2} &=\sum_{n=0}^\infty\left(\frac{-1}{27}\right)^ny^{3n+2}\\ &=y^2\sum_{n=0}^\infty\left(\frac{-1}{27}\right)^n(y^3)^n\\ \end{array} $
so, by the root test, this converges when $|y^3/27| < 1$ and diverges when $|y^3/27| > 1$. This is equivalent to $|y/3| < 1$ and $|y/3| > 1$ or $|x-3| < 3$ to converge and $|x-3| > 3$ to diverge.