$$\sum_{n=1}^{\infty}\frac{3^n-1}{3^{n+1}}$$
The answer to this is that it diverges but I have no idea how to take the limit of this.
It looks like if we direct sub in $n = \infty$ it will be $3^{\infty}$ which I have never learned.
Can anyone explain how to take the limit of this problem?
Well, I haven't really done sum problems other than geometric sums but I can do the limit... $$\lim_{x \to \infty}\frac{3^x-1}{3^{x+1}}=\lim_{x \to \infty}\frac{1-\frac{1}{3^x}}{\frac{3^{x+1}}{3^{x}}}$$ Divide top and bottom by $3^x$. This is the trick when solving limits which goes to infinity.
Note that $\frac{a^m}{a^n}=a^{m-n}$ so $\frac{3^{x+1}}{3^x}=3^{x+1-x}=3$ Now back to the other equation. $$\lim_{x \to \infty}\frac{1-\frac{1}{3^x}}{\frac{3^{x+1}}{3^{x}}}=\lim_{x \to \infty}\frac{1-\frac{1}{3^x}}{3}$$ $3^x$ goes to infinity so $\frac{1}{3^x}$ goes to 0. $$\lim_{x \to \infty}\frac{1-\frac{1}{3^x}}{3}=\frac{1-0}{3}$$ The sequence converges to $\frac{1}{3}$ but it's sum diverges as you know