Why does the continuous sum of the series (1/2^x) yield a smaller result than the discrete sum? The sum of the discrete series comes to be $1$:
$$S = \sum_{x=1}^{\infty}\frac{1}{2^{x}}$$
$$S = \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...$$
$$S = \frac{1}{2}+\frac{1}{2}\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...\right)$$
$$S=\frac{1}{2}+\frac{S}{2}$$
$$S=1$$
However, the sum of the similar continuous series comes out to be $0.721347520444$:
$$I=\int_{1}^{\infty}\frac{1}{2^{x}}dx$$
Taking $2^{-x}=t$, i.e., $x=1 => t=\frac{1}{2}$ and $x=\infty => t=0$:
$$-x=\log_{2}x$$
Differentiating with respect to x on both sides, we get:
$$dx=\frac{-1}{t\ln2}dt$$
Putting all this back into the original equation, we get:
$$I=\int_{\frac{1}{2}}^{0}t\cdot\frac{-1}{t\ln2}dt$$
$$=\int_{\frac{1}{2}}^{0}\frac{-1}{\ln2}dt$$
$$=\frac{1}{\ln2}\int_{\frac{1}{2}}^{0}-1dt$$
$$=\frac{1}{\ln2}\int_{0}^{\frac{1}{2}}dt$$
$$ =\frac{1}{2\ln2}$$
$$=0.721347520444$$
How is $I < S$, despite $I$ already having the values contained in $S$, and infinitely more?

Edit: Fixed typo
 A: The series you give is an upper Riemann sum for the integral, and hence is an overestimate.
A: An image of the things you're trying to compare.

Note that the blue lines are always above the curve.
The trick with summation vs integration is always that the summation assumes a certain width; integration takes a width that limits to 0.
A: Note that
$$\int_{1}^{\infty} 2^{-x}dx = \sum_{i=1}^{\infty} \int_{i}^{i+1} 2^{-x}dx$$
$$ = \sum_{i=1}^{\infty} \int_{1}^{2} 2^{-i}2^{-x}dx
=  \sum_{i=1}^{\infty} 2^{-i}\int_{1}^{2} 2^{-x}dx$$
So now it remains to show that $\int_{1}^{2} 2^{-x}dx$ is strictly less than $2^{-1}$. Note however that as $2^{-x}$ is a continous function that is no larger than $\frac{1}{2}=2^{-1}$ for all $x \in (1,2]$, and is as small as $\frac{1}{4}$ on the interval $(1,2]$ [namely at $x=2$], it follows that
$$\int_{1}^{2} 2^{-x}dx \ < \ \int_1^2 2^{-1}dx \ = \ 2^{-1} \times (2-1) \ = \ 2^{-1}.$$
A: Notice that $$\sum_{m=1}^{\infty}2^{-m}=\sum_{m=1}^{\infty}\int_m^{m+1}2^{-m}\,\mathrm{d}x=\sum_{m=1}^{\infty}\int_m^{m+1}2^{-\lfloor{x}\rfloor}\,\mathrm{d}x=\int_1^{\infty}2^{-\lfloor{x}\rfloor}\,\mathrm{d}x=\int_1^{\infty}2^{\lceil-x\rceil}\,\mathrm{d}x.$$ Since $\lceil-x\rceil\gt-x,$ it follows that $$2^{\lceil-x\rceil}\gt2^{-x},$$ and thus, $$\sum_{m=1}^{\infty}2^{-m}=\int_1^{\infty}2^{\lceil-x\rceil}\,\mathrm{d}x\gt\int_1^{\infty}2^{-x}\,\mathrm{d}x.$$ On the note that the integral includes the terms of the series already as part of the "continuous sum," as you call it, you are correct, but what you are ignoring is that those terms are being multiplied by quantities close to $0,$ and the same is true for all other "terms" that are part of this "continuous sum." Recall that the Riemann integral is the limit of a net of Riemann sums that look like $$\sum_{m=0}^n2^{-t_m}\Delta{x_m},$$ where $\Delta{x_m}\to0.$
