TITLE: Calculating the value of infinite limit of $2^{-n^2}/\sum_{k=n+1}^\infty 2^{-k^2}$
QUESTION [6 upvotes]: How to solve the limit?
                   $$\lim_{n \to \infty}\frac{2^{-n^2}}{\sum_{k=n+1}^\infty 2^{-k^2}}$$
My approach:-
               I have used logarithmic test to test the denominator sum for convergence as follows:-
$$ \lim_{ k\to\infty}k\log \frac{u_n}{u_{n+1}}=\lim_{k\to \infty}k\log\frac{2^{-k^2}}{2^{-(k+1)^2}}=\lim_{k\to\infty}(k+2k^2)\log2=\infty$$
Thus the infinite sum diverges.
The numerator term also diverges because it is a term from monotonically decreasing sequence which has no lower bound.
So overall solution is $$\infty$$
Is my attempt correct or wrong?

REPLY [7 votes]: In a different way,
$$\frac{2^{-n^2}}{\sum_{k=n+1}^\infty 2^{-k^2}}>\frac{2^{-n^2}}{\sum_{k=(n+1)^2}^\infty 2^{-k}}=\frac{2^{-n^2}}{2^{-(n+1)^2}\cdot2}=2^{-n^2+(n+1)^2-1}=2^{2n}$$
Therefore it diverges.

REPLY [5 votes]: Unfortunately, your attempt is seriously wrong. The sequence $2^{-n^2}$ is convergent, with limit zero ($0$ is a clear lower bound). Likewise, the sum $\sum_{k = n + 1}^{\infty} 2^{-n^2}$ is convergent by, say, comparison to a geometric series. In fact, both the numerator and denominator tend to zero. Finally, even if your first two conclusions were correct, your final conclusion wouldn't follow from them.

For a different approach, note that the denominator can be estimated by
$$\sum_{k = n + 1}^{\infty} 2^{-k^2} \approx 2^{-n^2 - 2n}$$ and so your sequence is bounded below by something like
$$\frac{2^{-n^2}}{2^{-n^2 - 2n}} = 2^{2n}$$