Induction Proof: $\sum_{k=1}^{n}\frac{a-1}{a^k}=1-\frac{1}{a^n}, a \ne 0$

Question

I'm having trouble showing equality for the $A(n+1)$ portion of the proof.

Prove by Induction: $$\sum_{k=1}^{n}\frac{a-1}{a^k}=1-\frac{1}{a^n}, a \ne 0$$

Base Case $(n=1)$: $$\frac{a-1}{a^1}=1-\frac{1}{a^1}$$ $$1-\frac{1}{a}=1-\frac{1}{a}$$ Therefore, $A(1)$ is true.

Assume $A(n+1)$ is true for some $n\ge1$. Then:

$$\sum_{k=1}^{n+1}\frac{a-1}{a^k}=1-\frac{1}{a^{n+1}}$$

Because of the lack of $n$ on both sides I am confused as to show we show $A(n+1)$ to be true. Any hints would be great.

Answer 1

Inductive step:

$$\sum_{k=1}^{n+1}\frac{a-1}{a^k}=\sum_{k=1}^{n}\frac{a-1}{a^k}+\frac{a-1}{a^{n+1}}=1-\frac{1}{a^{n}}+\frac{a-1}{a^{n+1}}=1-\frac{1}{a^{n+1}}$$

Answer 2

Hint:

The idea behind induction is that once the case $A(1)$ is proved you assume $A(k)$ is true for all $1\leq k\leq n$ to conclude that it musy be true for $A(n+1)$.

$$A(n+1) = \sum_{k=1}^{n+1} \frac{a-1}{a^k} = \sum_{k=1}^n \frac{a-1}{a^k} + \frac{a-1}{a^{n+1}} = 1-\frac{1}{a^n} + \frac{a}{a^{n+1}} - \frac{1}{a^{n+1}} = 1 - \frac{1}{a^{n+1}}$$

Answer 3

Based on the statement that $A(n)$ is true you must prove that $A(n+1)$ is true:$$\sum_{k=1}^{n+1}\frac{a-1}{a^{k}}=\sum_{k=1}^{n}\frac{a-1}{a^{k}}+\frac{a-1}{a^{n+1}}\stackrel{A\left(n\right)\text{ is true}}{=}1-\frac{1}{a^{n}}+\frac{a-1}{a^{n+1}}=1-\frac{1}{a^{n+1}}$$

Answer 4

An other way to prove your equality

You can do it with out induction proof. Indeed, $$\sum_{k=1}^n \frac{a-1}{a^k}=\sum_{k=1}^n\frac{1}{a^{k-1}}-\sum_{k=1}^n\frac{1}{a_k}=\sum_{k=0}^{n-1}\frac{1}{a^k}-\sum_{k=1}^n\frac{1}{a_k}=1-\frac{1}{a^n}+\sum_{k=1}^{n-1}\frac{1}{a^k}-\sum_{k=1}^{n-1}\frac{1}{a^k}=1-\frac{1}{a^n}$$