a definite integral identity? is it true? $\int_{a}^{b} f(x) dx $ i came across this identity , i don't know if it is true or not
$$\int_{a}^{b} f(x) \ \mathrm{d}x = (b-a) \sum_{n=1}^{\infty} \sum_{k=1}^{2^n - 1} \dfrac{(-1)^{k+1}}{2^{n}} f \left( a+ \left(\frac{b-a}{2^n}\right) k \right)$$
i tried to use the Riemann Sum but i couldn't proceed 
Anyone has an idea about it?
 A: Note that:
$$
\int_a^bf(x)dx=\lim_{N \to \infty}S_N \\
S_N=\sum_{i=1}^{2^N-1} \frac{b-a}{2^N}f(a+\frac{b-a}{2^N}i) =
\sum_{n=1}^N \sum_{k=1}^{2^n-1}\frac{b-a}{2^n}(-1)^{k+1}f(a+\frac{b-a}{2^n}k)
$$
Prove the last identity by induction:
For $N$=1, trivial.
For $N\to N+1$:
$$
\sum_{n=1}^{N+1} \sum_{k=1}^{2^n-1}\frac{b-a}{2^n}(-1)^{k+1} f(a+\frac{b-a}{2^n}k)\\
=\sum_{n=1}^{N} \sum_{k=1}^{2^n-1}\frac{b-a}{2^n}(-1)^{k+1} f(a+\frac{b-a}{2^n}k)
+\sum_{k=1}^{2^{N+1}-1}\frac{b-a}{2^{N+1}}(-1)^{k+1} f(a+\frac{b-a}{2^{N+1}}k)\\
=\sum_{i'=1}^{2^N-1} \frac{b-a}{2^N}f(a+\frac{b-a}{2^N}i')
+\sum_{k=1}^{2^{N+1}-1}\frac{b-a}{2^{N+1}}(-1)^{k+1} f(a+\frac{b-a}{2^{N+1}}k)\\
=\sum_{i'=1}^{2^N-1} \frac{b-a}{2^N}f(a+\frac{b-a}{2^N}i')
-\sum_{i'=1}^{2^{N}-1}\frac{b-a}{2^{N+1}} f(a+\frac{b-a}{2^{N+1}}(2i'))
+\sum_{i'=1}^{2^{N}}\frac{b-a}{2^{N+1}} f(a+\frac{b-a}{2^{N+1}}(2i'-1))\\
=\sum_{i'=1}^{2^N-1} \frac{b-a}{2^{N+1}} f(a+\frac{b-a}{2^{N+1}}(2i'))
+\sum_{i'=1}^{2^{N}}\frac{b-a}{2^{N+1}} f(a+\frac{b-a}{2^{N+1}}(2i'-1))\\
=\sum_{i=1}^{2^{N+1}-1} \frac{b-a}{2^{N+1}} f(a+\frac{b-a}{2^{N+1}}i)
$$
