find the value of the sumation?

Question

find the value of the limit $$\lim_{n\to\infty}\frac{2^{-n^2}}{ \sum_{k=n+1}^{\infty} \frac{1}{2^{k^2}}}=\infty$$ a) $0$

b) $some c ∈ (0,1)$

c) 1

d) ∞

this is the orginial questions

from my thinking the answer will be ∞

Answer 1

First we try to bound $\frac{2^{-n^2}}{ \sum_{k=n+1}^{\infty} \frac{1}{2^{-k^2}}}$, We know that:$$\frac{2^{-n^2}}{ \sum_{k=n+1}^{\infty} {2^{-k^2}}}=\frac{1}{ \sum_{k=n+1}^{\infty} {2^{n^2-k^2}}}=\frac{1}{2^{-2n-1}+2^{-4n-4}+2^{-6n-9}+...}>\frac{1}{2^{-2n}+2^{-4n}+2^{-6n}+...}=\frac{1}{(\frac{1}{4})^{n}+(\frac{1}{4})^{2n}+(\frac{1}{4})^{3n}+...}=\frac{1}{\frac{(\frac{1}{4})^{n}}{1-(\frac{1}{4})^{n}}}=4^n-1$$ then tending $n\to\infty $ leads us to: $$\lim_{n\to\infty}\frac{2^{-n^2}}{ \sum_{k=n+1}^{\infty} \frac{1}{2^{-k^2}}}=\infty$$