# Find the limit $\lim\limits_{n \to \infty}\frac{2^{-n^2}}{\sum\limits_{k=n+1}^{\infty} 2^{-k^2}}$

Find the limit of the following-$$\lim\limits_{n \to \infty}\frac{2^{-n^2}}{\sum\limits_{k=n+1}^{\infty} 2^{-k^2}}$$

My work:

We can see that the denominator is in geometric series, whose sum is $1$. So taking the limit we get $\infty?$ Am I right? But I think there is a problem in this reasoning as the powers are not starting from $0$ or $1$. Can we really consider a geometric series? Any help would be great. Thanks

• The denominator isn't a geometric series. – Peter Aug 10 '16 at 15:47
• @Peter I have the same confusion to be honest. So any hint? – Harry Potter Aug 10 '16 at 15:49

Hint : Multiply denominator and numerator with $$2^{n^2}$$
Note that what Peter hinted at gives: $$\frac{2^{-n^2}}{\sum_{k=n+1}^{\infty} 2^{-k^2}}=\frac{1}{\sum_{k=n+1}^\infty2^{-k^2+n^2}}$$ Now we know that $\sum_{n=1}^\infty2^{-n}=1$ . We should also note that the difference $n^2-(n+1)^2$ can be made arbitrarily large with big enough $n$. Hence, unrigorously, it appears that for any $N$ there exists a $n$ such that $$\sum_{k=n+1}^\infty2^{-k^2+n^2}\leq\sum_{i=N}^\infty2^{-i}.$$ From the convergence of the RHS we know that for every $\varepsilon>0$ there exists a $N_\varepsilon$ such that $$\sum_{i=N_\varepsilon}^\infty2^{-i}<\varepsilon.$$ From this we can conclude that the limit is infinity.
We have by integral test that$$\int_{n+1}^{\infty}\frac{1}{2^{t^{2}}}dt\leq\sum_{k=n+1}^{\infty}\frac{1}{2^{k^{2}}}\leq\frac{1}{2^{\left(n+1\right)^{2}}}+\int_{n+1}^{\infty}\frac{1}{2^{t^{2}}}dt$$ so let us now consider the integral, we have $$\int_{n+1}^{\infty}\frac{1}{2^{t^{2}}}dt=\int_{n+1}^{\infty}e^{-t^{2}\log\left(2\right)}dt=\frac{1}{\sqrt{\log\left(2\right)}}\int_{\left(n+1\right)\sqrt{\log\left(2\right)}}^{\infty}e^{-u^{2}}du$$ and since $$\frac{e^{-z^{2}}}{z+\sqrt{z^{2}+2}}\leq\int_{z}^{\infty}e^{-u^{2}}du\leq\frac{e^{-z^{2}}}{z+\sqrt{z^{2}+4/\pi}}$$ we get $$2^{n}\sum_{k=n+1}^{\infty}\frac{1}{2^{k^{2}}}\leq\frac{2^{n}}{2^{\left(n+1\right)^{2}}}+\frac{2^{n}}{\sqrt{\log\left(2\right)}}\frac{2^{-\left(n+1\right)^{2}}}{\left(n+1\right)\sqrt{\log\left(2\right)}+\sqrt{\left(n+1\right)^{2}\log\left(2\right)+4/\pi}}$$ and $$2^{n}\sum_{k=n+1}^{\infty}\frac{1}{2^{k^{2}}}\geq\frac{2^{n}}{\sqrt{\log\left(2\right)}}\frac{2^{-\left(n+1\right)^{2}}}{\left(n+1\right)\sqrt{\log\left(2\right)}+\sqrt{\left(n+1\right)^{2}\log\left(2\right)+2}}$$so finally we can conclude that $$\lim_{n\rightarrow\infty}\frac{2^{-n^{2}}}{\sum_{k=n+1}^{\infty}\frac{1}{2^{k^{2}}}}=\color{red}{\infty}.$$