Infinity Series to Approximate a fraction

Question

I have observed that the following series is a good approximation for $\frac{1}{10}$.

$\frac{1}{8}- \frac{1}{16} + \frac{1}{32} + \frac{1}{64} - \frac{1}{128} - \frac{1}{256} + \frac{1}{512} + \frac{1}{1024} - \frac{1}{2048} - \frac{1}{(2\cdot2048)} + \frac{1}{(4\cdot2048)} + \frac{1}{(8\cdot2048)} - \frac{1}{(16\cdot2048)} - \frac{1}{(32\cdot2048)} + ...$

I believe there is a pattern to the series. However, I cannot prove that pattern holds. Does it? Can it be proved? I am thinking some Taylor series might be the way to go but that is just a hunch.

Bob

Answer 1

The series is similar to $\sum \limits_{n=3}^{\infty} \frac{1}{2^n}=2-1/1-1/2-1/4=1/4$.

Note that $\sum \limits_{n=0}^{\infty} x^n=\frac{1}{1-x}$ for $|x|<1$.

EDIT, since I misread the question:

$$1/8 - 1/16 + 1/32 + 1/64 - 1/128 - 1/256 + 1/512 + 1/1024 - 1/2048 - 1/4096 + 1/8192 + 1/16384 - 1/32768 - 1 /65536 + ...=$$

$$\frac{1}{8}-\frac{1}{16}+\sum \limits_{n=2}^{\infty} \frac{(-1)^n}{2^{2n+1}}+\sum \limits_{n=2}^{\infty} \frac{(-1)^n}{2^{2n+2}}=$$

$$\frac{1}{8}-\frac{1}{16}+\frac{1}{40}+\frac{1}{80}=\frac{1}{10}$$

This expression can be evaluated using the formula given above

Answer 2

It is quite difficult to decipher the sign pattern you are using, especially at the start, but you might be able to rewrite your expression as

$$\frac18 - \frac1{16} + \frac12\left(1+\frac12-\frac14-\frac18\right)\left(\frac1{16}+\frac1{16^2}+\frac1{16^3}+\cdots\right)$$

which would then give $$=\frac1{16} + \frac12\times \frac98\times \frac1{15} = \frac1{10}$$

Answer 3

$$\frac{1}{8}-\frac{1}{16}+\sum \limits_{n=2}^{\infty} \frac{(-1)^n}{2^{2n+1}}+\sum \limits_{n=2}^{\infty} \frac{(-1)^n}{2^{2n+2}}=\dots$$