Think about what you are trying to prove, now you have an integral - draw what this means. The integral you have supplied is the area between the curve 1/x and teh x-axis bounded by x=2^n and x=2^n+1
So your 'lower' bound for the integral will be the area of the rectangle with width (2^n+1-2^n) and height of the lower of f(2^n) and f(2^n+1) where f(x)=1/x. Clearly the larger x is (x>0), the smaller 1/x=f(x) is, so your lower limit is the rectangle with height 1/(2^n+1) and the upepr limit will have height 1/(2^n).
A diagram will be very useful. So:
area of smaller rectangle < supplied integral < area of larger rectangle
area of smaller rectangle = height x width = 1/(2^n+1) x (2^n+1-2^n) = 1/(2^n+1) x (2-1)2^n = 1/(2^n+1) x 2^n = 2^n/2^n+1 = 1/2
area of larger rectangle = height x width = 1/2^n x 2^n (same width as above) = 1
therefore 2^-1 < supplied integral < 1
Q.E.D.