Show that the sequence $y_1,y_2,\cdots$ converges. My question is regarding the end of my proof: essentially, am I allowed to say "keep subtracting $1/2^i$ starting from i = 1 until we subtract at least $(n-1)/n$ from the infinite geometric series"? Or do I literally need to work with the formula that involves a finite number of partial sums in a geometric series. That seems very tedious. If this doesn't make sense, please read my proof below.
Proof: We want to show that $\{ y_i \}$ converges to a real number, and that's done iff it is a Cauchy sequence. Grab an arbitrary $1/n$ and suppose we have $y_b, y_a$ such that $b > a$. Then $|y_b-y_a| = |\sum_{i=1}^{b}x_i - \sum_{i=1}^{a}x_i| = |\sum_{i=a+1}^{b}x_i| \leq \sum_{i=a+1}^{b}|x_i| \leq \sum_{i=a+1}^{b}1/2^i$. To show $|y_b-y_a| \leq 1/n$, we need to pick an $m$ such that for $a,b \geq m, \sum_{i=a+1}^{b}1/2^i \leq 1/n$. Well, we can see that $\sum_{i=a+1}^{b}1/2^i \leq \sum_{i=a+1}^{\infty}1/2^i$, and so if we prove $\sum_{i=a+1}^{\infty}1/2^i \leq 1/n$, we are good. And so we basically need to find an $a$ that satisfies this, and let $m = a$. Well, we know that $1 = \sum_{i=1}^{\infty}1/2^i$, and so we subtract from $i = 1$ onwards until we subtract at least $(n-1)/n$. We know we can do this because $(n-1)/n$ is a finite number. Then, we let $a$ be the number of terms we subtracted plus 1, and then let $m=a$.
Yes, it's valid to say that since
$$1 = \sum_{i=1}^\infty \frac{1}{2^i},$$
and $0 < \frac{n-1}{n} < 1$, there is a natural number $m$ such that
$$ \sum_{i=1}^m \frac{1}{2^i} > \frac{n-1}{n} $$
and therefore
$$ \sum_{i=m+1}^\infty \frac{1}{2^i} < \frac{1}{n} $$
But an easier way to do this is with a change of summation variable, $j=i-a$ and $i=j+a$:
$$ \sum_{i=a+1}^\infty \frac{1}{2^i} = \sum_{j=1}^\infty \frac{1}{2^{a+j}} = \frac{1}{2^a} \sum_{j=1}^\infty \frac{1}{2^j} = \frac{1}{2^a} $$
Then just choose $a$ big enough that $\frac{1}{2^a} < \frac{1}{n}$.