Calculating the value of infinite limit of $2^{-n^2}/\sum_{k=n+1}^\infty 2^{-k^2}$

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.

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.

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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

$$\frac{2^{-n^2}}{2^{-n^2 - 2n}} = 2^{2n}$$