<s>[INST] Find $\lfloor 1000S \rfloor$ for $S =\sum_{n=1}^{\infty} \frac{1}{2^{n^2}} = \frac{1}{2^1}+\frac{1}{2^4}+\frac{1}{2^9}+\cdots.$ 
Let $$S = \displaystyle \sum_{n=1}^{\infty} \dfrac{1}{2^{n^2}} = \dfrac{1}{2^1}+\dfrac{1}{2^4}+\dfrac{1}{2^9}+\cdots.$$
  Find $\lfloor 1000S \rfloor$

My attempt was to recognize that consecutive perfect squares increase in the form $3,5,7,9,\ldots$ and be able to form a geometric series this way. I couldn't figure out how to make that work, though.
 [/INST] The sum of the first three terms is
$$1000\left[\frac{1}{2^{1}}+\frac{1}{2^{4}}+\frac{1}{2^{9}}\right] \simeq 564.453$$
The sum of the rest of the terms is bounded by (using a geometrical series)
$$1000\left[\frac{1}{2^{16}}+\frac{1}{2^{25}}+\frac{1}{2^{36}}+\ldots\right] < 1000\left(\frac{1}{2^{16}} + \frac{1}{2^{16+9}} + \frac{1}{2^{16+2\cdot 9}} + \ldots\right) \\= \frac{1000}{2^{16}}\frac{1}{1-\frac{1}{2^9}} < \frac{1000}{2^{15}} \simeq 0.031$$
From which we see that
$$564.453 < 1000S < 564.484 \implies \lfloor 1000S\rfloor = 564$$
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