Prove geometric sum with induction

Question

I'm not certain how to complete the proof:

Question:

Prove by induction that $1 − \frac{1}{3} + \frac{1}{9} − · · · + (\frac{−1}{3})^n = (\frac{3}{4})[1 − (\frac{−1}{3})^{n+1}]$, for every non negative integer $n$.

Solution

Base Step:

Verify that:

$LHS = 1 = (\frac{3}{4})[1 − (\frac{−1}{3})^{0+1}] = RHS$.

$RHS = (\frac{3}{4})[1 − (\frac{−1}{3})^1]\\ = (\frac{3}{4})[1 − (\frac{−1}{3})]\\ = (\frac{3}{4})(1 + \frac{1}{3})\\ = (\frac{3}{4})(\frac{4}{3}) = 1 = LHS.$

Inductive Step:

Assume that:

$1 − \frac{1}{3} + \frac{1}{9} − · · · + (\frac{−1}{3})^k = (\frac{3}{4})[1 − (\frac{−1}{3})^{k+1}]$, for some integer k.

We try to deduce that:

$1 − \frac{1}{3} + \frac{1}{9} − · · · + (\frac{−1}{3})^{k+1} = (\frac{3}{4})[1 − (\frac{−1}{3})^{k+2} ]$.

$LHS = 1 − \frac{1}{3} + \frac{1}{9} − · · · + (\frac{−1}{3})^{k+1} \\ = 1 − \frac{1}{3} + \frac{1}{9} − · · · + (\frac{−1}{3})^{k} + (\frac{−1}{3})^{k+1}\\ = \frac{3}{4}[1-(\frac{-1}{3})^{k+1}] + (\frac{-1}{3})^{k+1}\\ ... $

Lost from this point onwards.

Answer 1

You just about have it. From your last line, you get

$$\begin{equation}\begin{aligned} \frac{3}{4}\left[1-\left(\frac{-1}{3}\right)^{k+1}\right] + \left(\frac{-1}{3}\right)^{k+1} & = \frac{3}{4} - \frac{3}{4}\left(\frac{-1}{3}\right)^{k+1} + \left(\frac{-1}{3}\right)^{k+1} \\ & = \frac{3}{4} + \frac{1}{4}\left(\frac{-1}{3}\right)^{k+1} \\ & = \frac{3}{4} + \frac{(-3)}{4}\left(\frac{1}{-3}\right)^{k+2} \\ & = \frac{3}{4}\left[1 - \left(\frac{-1}{3}\right)^{k+2}\right] \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

which is the RHS of what you are trying to deduce, thus completing your induction procedure.

Answer 2

So far so good. Let $q = -\frac{1}{3}$. Thus, following your lines, you have shown $$ \sum_{j=1}^{k+1} q^j = \sum_{j=1}^{k} q^j + q^{k+1} = \frac{1-q^{k+1}}{1-q} + q^{k+1}.$$ To finish the proof, note that $$q^{k+1} = \frac{1-q}{1-q} q^{k+1}.$$

Answer 3

After a small amount of work, my final solution is:

\begin{align} LHS &= 1 − \frac{1}{3} + \frac{1}{9} − · · · + (\frac{−1}{3})^{k+1} \\ &= 1 − \frac{1}{3} + \frac{1}{9} − · · · + (\frac{−1}{3})^{k} + (\frac{−1}{3})^{k+1}\\ &= \frac{3}{4}[1-(\frac{-1}{3})^{k+1}] + (\frac{-1}{3})^{k+1}\\ &= \frac{3}{4}[1-(\frac{-1}{3})^{k+1}] + \frac{3}{4}[\frac{4}{3}(\frac{-1}{3})^{k+1}]\\ &= \frac{3}{4}[1-(\frac{-1}{3})^{k+1} + \frac{4}{3}(\frac{-1}{3})^{k+1}]\\ &= \frac{3}{4}[1+(\frac{4}{3}-1)(\frac{-1}{3})^{k+1}]\\ &= \frac{3}{4}[1 - (\frac{-1}{3})^{k+2}]\\ &= RHS \end{align}