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?

$$1+\frac12+\frac{1}{2^2}+\cdots+\frac{1}{2^n}=1+\frac{1}{2} \cdot \frac{\left(\frac{1}{2}\right)^{n}-1}{\frac{1}{2}-1}.$$ Now it is obviously, that it is bounded.
