$\sum_{i=1}^n {2\over3^i}={2\over3}+{2\over9}+\dots+{2\over3^n}=1-{({1\over3})^n}$
I had this problem in class and we proved using 2 different methods: contradiction and mathematical induction. I thought it was understood, I just got bumped into certain point.
Please point it out which step I'm thinking wrong.
For the contradiction, We assume that there is some integer n for which $i=1$ is false. And we are applying smaller positive integer smaller than 1.
for the smallest n, ${2\over3}+{2\over9}+{\dots}+{1\over3^{n-1}}$ indicates that our assumption $i=1$ is false.
(I don't remember how the calculation was made for this proof by contradiction.)
Therefore, our assumption was true.
For induction,
Try out the base case with applying $i=1$ inductive hypothesis would be ${2\over3}=1-{1\over3}$
What would be the next step?
By contradiction.
Assume that $\sum_{i=1}^{n}\frac{2}{3^{i}}\neq1-\frac{1}{3^{n}}$ for some $n\in\mathbb{N}$ and let $k$ be the smallest positive integer with $\sum_{i=1}^{k}\frac{2}{3^{i}}\neq1-\frac{1}{3^{k}}$.
Then evidently $k>1$.
The minimality of $k$ implies that $\sum_{i=1}^{k-1}\frac{2}{3^{i}}=1-\frac{1}{3^{k-1}}$ so that: $$\sum_{i=1}^{k}\frac{2}{3^{i}}=\sum_{i=1}^{k-1}\frac{2}{3^{i}}+\frac{2}{3^{k}}=1-\frac{1}{3^{k-1}}+\frac{2}{3^{k}}=1-\frac{1}{3^{k}}$$
A contradiction has been found and we conclude that $\sum_{i=1}^{n}\frac{2}{3^{i}}=1-\frac{1}{3^{n}}$ for any $n\in\mathbb{N}$.