How to prove integral(1 to infinity) (1/2^x) dx < infinity

Prove
integral(1 to infinity) (1/2^x) dx < infinity

Lower limit is 1, upper limit is inifnity.

∫ 1/2^x dx
= - 2^-x / ln2 + c
Plugging limits x = 1 to ∞,
= - (1/ln2) * lim (x → ∞) [2^-x - 2^-1]
= 1/(2ln2) < ∞.

integral(1 to infinity) (1/2^x) dx
< Sum (1 to infinity) (1/2^n) (curve is below the series of rectangles)
=1/2 + 1/4 + 1/8+....
=1

integral 2^-x dx from 1 to infinite= -2^-x/ln2 from infinite to 1,= 0+1/(2ln2)