What is the limit of $\sum^{i=n}_{i=0} \left(-\frac{2}{3}\right)^i$? What is the limit of $$\sum^{i=n}_{i=0} \left(-\frac{2}{3}\right)^i$$ ?
I am currently taking a introductory course in real analysis. Hints will be appreciated!

As a soft question, can I ask what should I search for in the search bar to find my answer? I am pretty sure there are similar sort of questions asked here before, and I want to try avoiding duplicating questions as much as possible.


Let $|r|<1$ and let $S=\displaystyle\sum_{i=0}^n r$.
Note that $rS=r+r^2+\ldots+r^{n+1}$, so that $rS-S=r^{n+1}-1$. It follows that
$$S=\frac{1-r^{n+1}}{1-r}.$$
Since $|r|<1$, $\lim_{n\to\infty}r^{n+1}=0$, so that the limit of the sum becomes $\frac{1}{1-r}$. In your case, $r=-\frac{2}{3}$, so the limit is $\frac{1}{1+\frac{2}.{3}}=\frac{3}{5}$
