negative exponent problem $$\sqrt{\frac{1}{3^0 + 3^{-1} + 3^{-2} + 3^{-3} + 3^{-4}}}$$
Does this equal =
$$
\begin{align*}
& \sqrt{3^0 + 3^1 + 3^2 + 3^3 + 3^4} \\
=&\sqrt{1 + 3 + 9 + 27 + 81} \\
=&\sqrt{121} \\
=&11. 
\end{align*}
$$
The answer is apparently $\frac{9}{11}$ and I'm not sure what rule of negative exponents I got wrong.
The rule I'm using, incorrectly, is this:
$$\frac{1}{3^{-2}} = 3^2 = 9.$$

$$\sqrt{\frac1{3^0 + 3^{-1} + 3^{-2} + 3^{-3} + 3^{-4}}}=\sqrt{\frac{3^4}{3^4(3^0 + 3^{-1} + 3^{-2} + 3^{-3} + 3^{-4})}}$$
$$=\sqrt{\frac{3^4}{3^4+3^3+3^2+3^1+3^0}}=\sqrt{\frac{81}{121}}=\frac9{11}$$
