I don't know the answer to this infinite sum due to it being "slightly unusual". We "apply" the Thue-Morse sequence (e.g.: $ABBABAAB\ BAABABBA\ BAABABBA\ ABBABAAB$) to the signs of the terms in the infinite(ly long) sum $\sum_{n=0}^{\infty}\frac{1}{2^n}$, and when written out the sum becomes: $$+\frac{1}{1}-\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}+\frac{1}{64}-\frac{1}{128}...$$ I thought of separating it out into $+$'s and $-$'s but that didn't seem to get me anywhere and I can't see any patterns in the partial sums. (Note that I am not very good with infinite sums.) What I'm asking for is the answer to this sum along with how the answer is achieved and whether it relates to the numbers $e$ and $\pi$ (by "relates to" I mean whether there is anything you can do to $e$ and $\pi$ in order to turn them into the answer to this infinite sum, I'm not sure what the term for this is though). I am reasonably sure that this series converges because $\sum_{n=0}^{\infty}\frac{1}{2^n}=2$ and $\sum_{n=0}^{\infty}-\frac{1}{2^n}=-2$ which implies that it would be between +2 and −2, but I do not know if this is valid reasoning.
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1You are sure the series is convergent and does not diverge? – Soham Apr 03 '16 at 14:32
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1@tatan I am reasonably sure because $\sum_{n=0}^{\infty}\frac{1}{2^n}=2$ and $\sum_{n=0}^{\infty}-\frac{1}{2^n}=-2$ which implies that it would be between $+2$ and $-2$. – god of llamas Apr 03 '16 at 14:35
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1The sequence is known as the Thue-Morse sequence. – Catalin Zara Apr 03 '16 at 14:41
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@CatalinZara Thanks, I'll put it in the answer. – god of llamas Apr 03 '16 at 14:42
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This is the Thue-Morse sequence (A010060 in OEIS). For programmers, it is the bit parity of the non-negative integers: $s_n = A$ if $n$ has an even number of $1$'s in its binary expansion. – TonyK Apr 03 '16 at 14:43
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@TonyK: It was not the Thue-Morse sequence when I wrote the comment. I looked for a plausible continuation. But now that it is the Thue-Morse sequence, I will delete the comment. – André Nicolas Apr 03 '16 at 14:50
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1@tatan: The series is absolutely convergent, hence convergent. :) – Andrew D. Hwang Apr 03 '16 at 14:51
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6The series converges and is equal to 2-4P=0.35018..., where P is the Thue-Morse constant. – pregunton Apr 03 '16 at 14:57
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There is an almost similar constant, the Prouhet-Thue-Morse constant, which consists in the real number whose binary expansion is the Thue-Morse sequence and has been proven to be transcendental: $$\tau = \sum_{i=0}^{\infty} \frac{t_i}{2^{i+1}}$$ where $t_i$ is the $i$-th term of the Thue-Morse sequence: $(t_0, t_1, t_2, t_3, \dots) = (0, 1, 1, 0, \dots)$.
In your case, you are considering $$S = \sum_{i=0}^{\infty} \frac{(-1)^{t_i}}{2^{i}} = \sum_{i=0}^{\infty} \frac{1 - 2t_i}{2^{i}} = \sum_{i=0}^{\infty} \frac{1}{2^{i}} - 4\sum_{i=0}^{\infty} \frac{t_i}{2^{i+1}}$$
Therefore, $S = 2 - 4\tau = 0.35018386544\dots$, which is a transcendental number (since $\tau$ is).
yoann
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After a bit of staring at how you "transformed" (for lack of a better term) $\sum_{i=0}^{\infty} \frac{(-1)^{t_i}}{2^{i}}$ into $\sum_{i=0}^{\infty} \frac{1}{2^{i}} - 4\sum_{i=0}^{\infty} \frac{t_i}{2^{i+1}}$, I understood it :D – god of llamas Apr 03 '16 at 15:20