Find the value of $\lim\limits_{n\rightarrow \infty}\left( \dfrac{2^{-n^2}}{\sum\limits_{k=n+1}^{\infty}2^{-k^2}}\right)=$?

Find the value of $ \lim\limits_{n\rightarrow \infty}\left( \dfrac{2^{-n^2}}{\sum\limits_{k=n+1}^{\infty}2^{-k^2}}\right)=$ ?

My answer:

Take $ r_{n+1} = \sum\limits_{k=n+1}^{\infty}2^{-k^2}$ .

If $ \lim\limits_{n\rightarrow \infty}\left( \dfrac{2^{-n^2}}{\sum\limits_{k=n+1}^{\infty}2^{-k^2}}\right)$ exists, then \begin{align*} 1+\lim\limits_{n\rightarrow \infty}\left( \dfrac{2^{-n^2}}{\sum\limits_{k=n+1}^{\infty}2^{-k^2}}\right)&=\lim\limits_{n\rightarrow \infty}\left( 1+\dfrac{2^{-n^2}}{\sum\limits_{k=n+1}^{\infty}2^{-k^2}}\right)\ &=\lim\limits_{n\rightarrow \infty}\left( \dfrac{\sum\limits_{k=n}^{\infty}2^{-k^2}}{\sum\limits_{k=n+1}^{\infty}2^{-k^2}}\right)\ &=\lim\limits_{n\rightarrow \infty}\dfrac{r_n}{r_{n+1}}=?? \end{align*} Moreover, I know $ r_n$ is a remainder of the series $ \sum\limits_{k=1}^{\infty}2^{-k^2}$ , which is convergent.$ ~~~~\left(\mbox{ By ratio test,} ~~\lim\limits_{n\rightarrow\infty} \dfrac{2^{-(n+1)^2}}{2^{-n^2}}=\lim\limits_{n\rightarrow\infty}\dfrac{1}{2.2^{2n}}=0<1 \right)$