Exactly expressing integral as a sum Apparently (i.e. according to my professor), the following holds:$$\int_a^b f(x) dx = (b-a)\sum_{n=1}^\infty \sum_{m=1}^{2^n-1} (-1)^{m+1}2^{-n}f(a+m(b-a)2^{-n}).$$How would one go about proving such a formula?
 A: We can break an alternating sum into the difference of the non-alternating sum and twice the sum of the even terms:
$$
\begin{align}
&\sum_{m=1}^{2^n}(-1)^{m-1}(b-a)2^{-n}f(a+m(b-a)2^{-n})\\
&=\sum_{m=1}^{2^n}(b-a)2^{-n}f(a+m(b-a)2^{-n})
-2\sum_{m=1}^{2^{n-1}}(b-a)2^{-n}f(a+2m(b-a)2^{-n})\\
&=\underbrace{\sum_{m=1}^{2^n}(b-a)2^{-n}f(a+m(b-a)2^{-n})}_{\text{Riemann Sum with $2^n$ partitions}}
-\underbrace{\sum_{m=1}^{2^{n-1}}(b-a)2^{-n+1}f(a+m(b-a)2^{-n+1})}_{\text{Riemann Sum with $2^{n-1}$ partition}}\tag{1}
\end{align}
$$
The sum of $(1)$ telescopes:
$$
\begin{align}
&\sum_{n=1}^\infty\sum_{m=1}^{2^n}(-1)^{m-1}(b-a)2^{-n}f(a+m(b-a)2^{-n})\\
&=\int_a^bf(x)\,\mathrm{d}x-(b-a)f(b)\tag{2}
\end{align}
$$
The terms of $(2)$ where $m=2^n$ sum to
$$
-(b-a)\sum_{n=1}^\infty2^{-n}f(b)=-(b-a)f(b)\tag{3}
$$
Subtracting $(3)$ from $(2)$ yields
$$
\sum_{n=1}^\infty\sum_{m=1}^{2^n-1}(-1)^{m-1}(b-a)2^{-n}f(a+m(b-a)2^{-n})
=\int_a^bf(x)\,\mathrm{d}x\tag{4}
$$
A: Let $l = b-a$, now put multiplier $l$ inside the inner series and let's consider members of outer series for different $n$.
$n = 1:$
$$\sum_{m=1}^{1} (-1)^{m+1}\frac l2 f(a+\frac{lm}{2}) = \frac l2f(a + \frac l2)$$
$n = 2:$
$$\sum_{m=1}^{3} (-1)^{m+1}\frac l4 f(a+\frac{lm}{4}) = \frac l4f(a + \frac l4) - \frac l4f(a + \frac {2l}{4}) + \frac l4f(a + \frac {3l}{4})$$
Now consider partiam sums $S_n$ of the outer series.
$$S_1 = \frac l2 f(a+\frac l2)$$
$$S_2 = \frac l2 f(a+\frac l2) + \frac l4f(a + \frac l4) - \frac l4f(a + \frac {2l}{4}) + \frac l4f(a + \frac {3l}{4}) = \frac l4f(a + \frac l4) + \frac l4f(a + \frac {2l}{4}) + \frac l4f(a + \frac {3l}{4})$$
From this we can see the form of $S_n$:
$$S_n = \sum_{k=1}^{2^n-1}\frac{l}{2^n}f(a + \frac{kl}{2^n})$$
You may plot a graph to understand this expression better and I will explain it analytically: for every $n$ we split $[a,b]$ into $2^n$ parts and count sum of values of $f(x)$ in the start of each segment multiplied by the segment length. This means that this expression is a particular case of integral partial sum $\sigma_n = \sum_{k=1}^{n-1}f(\xi_k)\Delta x_k$, where $x_k$ are some points from $[a,b]$ and $\xi_k \in [x_k, x_{k+1}]$. In our case $\Delta x_k$ is independent on k and equals $\frac{l}{2^n}$ and $\xi_k = x_k$.
As
$$\lim_{n \to \infty}\sigma_n = \int_a^bf(x)dx$$
and
$$\lim_{n \to \infty}S_n = \sum_{n=1}^{\infty}\sum_{m=1}^{2^n-1} (-1)^{m+1}\frac {l}{2^n} f(a+\frac{lm}{2^n}),$$
we have proved that the initial series converges to $\int_a^bf(x)dx$.
