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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 $$ \sum_{n=1}^\infty \left|(X_n)^n\right|\le 2$$

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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.

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    Ahhh so it has to start at n = 0! Just to be sure $$ \sum_{n=0}^\infty q^n$$ if |q|<1 I can say that $$ \frac {1}{1-q} = Sn$$ right? but why have I sometime seen $$ Sn= \frac {q}{1-q}$$ for geometric series is it when we start at n=1 instead of n= 0? – 2017-01-11
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    Yes, ok, I assume that with $S_n$ you denote the infinite sum, but yes that is it. – 2017-01-11
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    so to be clear the following is true? (for |q| <1) $$\sum_{n=0}^\infty q^n = \frac{1}{1-q}$$ or $$\sum_{n=1}^\infty q^n = \frac{q}{1-q}$$ – 2017-01-11
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    Yes, both versions are correct. The first one is the Geometric Series (starting from $n=0$), the second one can be derived directly from the first one, since $q^0=1$, hence $$\frac{1}{1-q}-1=\frac{1-1+q}{1-q}=\frac{1}{1-q}$$ – 2017-01-11
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$\sum_{n=1}^\infty ({\frac{1}{2}})^n=\sum_{n=0}^\infty ({\frac{1}{2}})^n-1=2-1=1$

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    The downvote is not justified. This answer directly addresses the problem in th OP's calculations. So, +1 from me. – 2017-01-11