Prove that $\frac23 + \frac29 + \dots + \frac{2}{3^n} = 1 - \left( \frac13 \right) ^n$

Question

Prove that: $$\frac23 + \frac29 + \dots + \frac{2}{3^n} = 1 - \left( \frac13 \right) ^n ,\text{for all } n > 0$$

I'm trying to do it using induction, thus:

For the smallest possible $n$, we have $n = 1 \Rightarrow \frac23 = 1 - \left( \frac13 \right) ^1 = \frac23$

Then $n-1 > 0$ and it should hold that:

$$\frac23 + \frac29 + \dots + \frac{2}{3^{n-1} } = 1 - \left( \frac13 \right) ^{n-1}$$

if we add $\frac{2}{3^n}$ on both sides we have:

$$\frac23 + \frac29 + \dots + \frac{2}{3^{n-1} } + \frac{2}{3^n} = 1 - \left( \frac13 \right) ^{n-1} + \frac{2}{3^n}$$

Now, here is my concern:

Simplifying the r.h.s. we have: $$1 - \left( \frac13 \right) ^{n-1} + \frac{2}{3^n} = 1 - 5\left( \frac13 \right) ^{n}$$

i.e. there is a factor of $5$ difference from the target? What am I doing wrong? What should be the inductive conclusion?

Answer 1

$$ 1-\frac{1}{3^{n-1}}+\frac{2}{3^n} = 1-\frac{3}{3^n}+\frac{2}{3^n}=1-\frac{1}{3^n} $$

Answer 2

It's a GP with r equals 1/3.. Substitute it in the formula you get the answer.

Besides your calculations are wrong

Answer 3

try prove without induction :

Let $\frac 2 3 + ... +\frac 2 {3^n} = A$

$\frac 1 3 (2+ \frac 2 3 +...+ \frac 2 {3^(n-1)}) = A $

$ 3A = 2 + ( A - \frac 2 {3^n} ) $

$ 2A = 2 - \frac 2 {3^n}$

and divide 2 to both side of equation