Manipulating Series To Find Maximum Interval Of Convergence Original Problem: Use the formula for the sum of a geometric series to find a power series that converges to expressions. For what values of y does the series converge?
$$8\over 4+y$$
Formula for the sum of a geometric series are written in the form $$a\over1-r$$ where a is the first term and r is the common ratio.
I did various approaches to this problem and got different interval of convergence, however, I wanted to find the maximum interval of convergence, so I tried this:
$$\lim_{b \to \infty } {{1\over b}\over {1\over b}}{8\over4+y}$$
$$\lim_{b \to \infty } {{8\over b} \over {{4+y}\over b}}$$
From this, I added a fancy 0 to the denominator, and then I rewrote the expression in terms of the sum of a geometric series.
$$\lim_{b \to \infty } {{8\over b} \over 1-1+ {{4+y}\over b}}$$
$$\lim_{b \to \infty } {{8\over b} \over 1-(1- {{4+y}\over b})}$$
Last, I transform the expression into the Riemann sums, so I could use the Ratio Test to find the interval of convergence (1*). 
$$\lim_{b \to \infty } \sum_{n=0}^\infty {8 \over b} \left( {1- {{4+y}\over b}} \right)^n$$
Ratio Test:
$$\lim_{b \to \infty }\lim_{n \to \infty } \left\lvert{{{8 \over b}({1- {{4+y}\over b}})^{n+1}}\over{{{8 \over b}({1- {{4+y}\over b}})^n}}} \right\rvert < 1 $$
This simplifies to:
$$\lim_{b \to \infty } \left(-1 < 1-{{4+y}\over b} < 1\right) $$
$$\lim_{b \to \infty} (2b-4 > y > -4) $$
Which I then simplify to:
$$\infty > y > -4 $$
My main question here, am I able to do this to find an interval of convergence? My concern is that when I pick a value for y that was in my interval and put the Riemann sum (at 1*) into my calculator, TI-Nspire CX, it wasn't able to solve and instead it spit it right back. My second question was now that I have this huge interval of convergence because my calculator can't calculate the sum, can I just use the original expression $8 / (4+y)$ to evaluate a sum given a chosen y that's in the interval?
 A: \begin{align*}
\frac{8}{4+y} &=
\frac{8}{4+y}\cdot 
\frac{\frac{1}{4}}{\frac{1}{4}}\\[4pt]
&=\frac{2}{1+\frac{y}{4}}\\[6pt]
&=\frac{2}{1-\left(-\frac{y}{4}\right)}\\[4pt]
\end{align*}
which can be expressed as a geometric series with $$a=2,\;r = -\frac{y}{4}$$
hence converges if and only if $\large{\left|-\frac{y}{4}\right|<1}$, so the interval of convergence is $-4 < y < 4$.

But in your comments, you asked whether the interval for $y$ can be enlarged by shifting the center.

Let's say you shift the center to $y=1$. Thus

\begin{align*}
\frac{8}{4+y} &=
\frac{8}{5+(y-1)}\\[4pt]
&=\frac{8}{5+(y-1)}\\[4pt]
&=\frac{8}{5+(y-1)}\cdot \frac{\frac{1}{5}}{\frac{1}{5}}\\[4pt]
&=\frac{\frac{8}{5}}{1+\frac{y-1}{5}}\\[6pt]
&=\frac{\frac{8}{5}}{1-\left(-\frac{y-1}{5}\right)}\\[4pt]
\end{align*}
which can be expressed as a geometric series with $$a=\frac{8}{5},\;r = -\frac{y-1}{5}$$
hence converges if and only if $\large{\left|-\frac{y-1}{5}\right|<1}$, so the interval of convergence is $-4 < y < 6$.

In general, if you shift the center to $y=c$, where $c \in \mathbb{R}$ and $c \ne -4$, the new radius of convergence will be $|4+c|$, and the interval of convergence for $y$ will the open interval centered at $y=c$, with one endpoint at $-4$.

So yes, the interval of convergence for $y$ can be made larger by centering the series further away from $-4$, but note, regardless of the choice of center, the interval of convergence will still be bounded.
