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

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.$~~~~\left(\mbox{ By ratio test,} ~~\lim\limits_{n\rightarrow\infty} \dfrac{2^{-(n+1)^2}}{2^{-n^2}}=\lim\limits_{n\rightarrow\infty}\dfrac{1}{2.2^{2n}}=0<1 \right)$

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  • $\begingroup$ both $r_n\to 0, r_{n+1}\to 0$, so it is indeterminate form of limit $\endgroup$ – farruhota May 10 at 6:21
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For $k \geq n+1$ it is easy to see that $n^{2}-k^{2} \leq -k$. Hence $\sum\limits_{k=n+1}^{\infty} 2^{n^{2}-k^{2}} \leq \sum\limits_{k=n+1}^{\infty} 2^{-k} \to 0$. Hence the required limit is $\infty$.

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