$\lim\limits_{i\mapsto \infty} \sum_{i=1}^\infty \sin(\theta_{i+1}-\theta_{i})$ = ? , where $\theta_{i}= \pi\sum_{j=0}^i \frac{1}{{(2)}^j}$

Question

$\lim\limits_{i\mapsto \infty} \sum_{i=1}^\infty \sin(\theta_{i+1}-\theta_{i})$ = ?

$\theta_{i}= \pi\sum_{j=0}^i \frac{1}{{(2)}^j} = \pi \left(2 - \frac{1}{2^{i}}\right)$

The limit $\lim\limits_{i\mapsto \infty} \sum_{i=1}^\infty \sin(\theta_{i+1}-\theta_i)$ exist and is not difficult to show that but what is the value of the limit?

using Taylor's series and some manipulations, I obtained $\lim\limits_{i\mapsto \infty} \sum_{i=1}^\infty \sin(\theta_{i+1}-\theta_{i})=\sum_{k=0}^\infty \frac {(-1)^k\pi^{2k+1}}{(2k+1)!2^{2k+1}(2^{2k+1}-1)}$

The series converges very rapidly and summing up to $k=2$ I got $\sum_{k=0}^2 \frac {(-1)^k\pi^{2k+1}}{(2k+1)!2^{2k+1}(2^{2k+1}-1)}=1.4810865$

Answer 1

using Taylor's series and some manipulations, I obtained $\lim\limits_{i\mapsto \infty} \sum_{i=1}^\infty \sin(\theta_{i+1}-\theta_{i})=\sum_{k=0}^\infty \frac {(-1)^k\pi^{2k+1}}{(2k+1)!2^{2k+1}(2^{2k+1}-1)}$

The series converges very rapidly and summing up to $k=2$ I got $\sum_{k=0}^2 \frac {(-1)^k\pi^{2k+1}}{(2k+1)!2^{2k+1}(2^{2k+1}-1)}=1.4810865$

it turns out $1.4810865 - 0.000082 \le \lim\limits_{i\mapsto \infty} \sum_{i=1}^\infty \sin(\theta_{i+1}-\theta_{i}) \le 1.4810865 + 0.000082$