Find $\lfloor 1000S \rfloor$ for $S =\sum_{n=1}^{\infty} \frac{1}{2^{n^2}} = \frac{1}{2^1}+\frac{1}{2^4}+\frac{1}{2^9}+\cdots.$

Question

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.

The key is to realize that the terms decay so fast than we only need a few terms to evaluate it to $3$ decimal digits (which is what $\lfloor 1000S\rfloor$ means). — Winther, Jan 10 '16 at 00:14

score 10 · Accepted Answer · answered Jan 10 '16 at 00:32

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

$$564.453 < 1000S < 564.484 \implies \lfloor 1000S\rfloor = 564$$

Find $\lfloor 1000S \rfloor$ for $S =\sum_{n=1}^{\infty} \frac{1}{2^{n^2}} = \frac{1}{2^1}+\frac{1}{2^4}+\frac{1}{2^9}+\cdots.$

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