Geometric series convergence and upper bound of its absolute value

Question

If I know the following: $1/4 \le |X_n| \le 1/2$ for all $n\ge 1$, then I must prove that $$\sum_{n=1}^\infty (X_n)^n$$ converges and that $$ \left|\sum_{n=1}^\infty (X_n)^n\right|\le 1$$

I used comparaison trick to show that $$\sum_{n=1}^\infty \left({\frac{1}{2}}\right)^n$$ converges to ${\frac{1}{1-{\frac{1}{2}}}}=2$, thus $\sum_{n=1}^\infty \left|(X_n)^n\right|$ also converges, since $|X_n|^n \le \left(\frac{1}{2}\right)^n$ so $$ \sum_{n=1}^\infty \left|(X_n)^n\right|\le 2$$

Answer 1

Hint: You start from $n=1$ but the geometric series starts from $n=0$. So, you must substract $X^0_n=1$ from the upper bound that you found.

Answer 2

$\sum_{n=1}^\infty ({\frac{1}{2}})^n=\sum_{n=0}^\infty ({\frac{1}{2}})^n-1=2-1=1$