Is $X_n = 1+\frac12+\frac{1}{2^2}+\cdots+\frac{1}{2^n}; \forall n \ge 0$ bounded? Is $X_n = 1+\frac12+\frac{1}{2^2}+\cdots+\frac{1}{2^n}; \forall n \ge 0$ bounded?
I have to find an upper bound for $X_n$ and i cant figure it out, a lower bound can be 0 or 1 but does it have an upper bound?

$1+\frac12+\frac{1}{2^2}+\cdots+\frac{1}{2^n}=1+\frac{\frac{1}{2}((\frac{1}{2})^n-1)}{(\frac{1}{2})-1}=1+1-(\frac{1}{2})^n\leq2$ $\forall$ $n\geq0.$
