Prove if this converges:
$$\sum_{n=1}^\infty \frac{2^n+3}{3^n-1}$$
pf: using geometric
$$0 < \frac{2^n+3}{3^n-1} \leq \frac{2^n + 2 \times 2^n}{3^n-\frac{3^n}{2}} = \cdots $$ and so on
I know how to do the rest but my question is that where in the world did my teacher get
$$\frac{2^n + 2 \times 2^n}{3^n-\frac{3^n}{2}}$$
A slightly better bound would be $$\frac{2^n+3}{3^n-1}=\frac{2^n(1+3\cdot 2^{-n})}{3^n(1-3^{-n})}\leq \frac{5/2}{2/3}\frac{2^n}{3^n}$$
You teacher increased the numerator (from $2^n+3$ to $2^n+ 2\cdot 2^n$, since $3\leq 2\cdot 2^n$), and decreased the denominator (from $3^n-1$ to $3^n-3^n/2$, since $1<3^n/2$), thus obtains a fraction (i.e. $\frac{2^n+2\cdot 2^n}{3^n-3^n/2}$) which is bigger than the original (i.e. $\frac{2^n+3}{3^n-1}$).
Ratio test: $$ \frac{\left(\dfrac{2^{n+1}+3}{3^{n+1}-1}\right)}{\left(\dfrac{2^n+3}{3^n-1}\right)} \to \frac 2 3 \text{ as }n\to\infty, $$ so the series converges.
