Prove by induction that $(1-a)^n ≥ 1-na$, $∀ n≥1$ for appropriate $a$.

Question

Prove by induction that $(1-a)^n ≥ 1-na$, $∀ n≥1$ for appropriate $a$.

Okay, so I have no problem with this except the requirements on $a$ for this inequality to hold. My lecturer claims we require $0<a<1$ but I can't see the need for the $a>0$ condition. I think it works for all $a<1$.

In the proof I multiplied the inequality by $1-a$ requiring $1-a>0$ and thus $a<1$ to preserve the inequality. I can't think why any of the steps would require a to be positive or non-zero. Any thoughts?

Here is my proof: Clearly the inequality holds for $n=1$ since $(1-a)^1=1-(1)a$

Now if we assume $(1-a)^k ≥ 1-ka$ for $k≥1$ and consider the case for $k+1$:

$(1-a)^{k+1}=(1-a)(1-a)^k ≥ (1-a)(1-ka)$ by assumption when $1-a>0$ so $a<1$.

So $(1-a)$^$(k+1)$ $≥ 1-ka-a+ka^2 = 1-(k+1)a +ka^2 ≥ 1-(k+1)a$.

So it holds for the $k+1$ case, so the inequality holds for all $n≥1$ by induction when $a<1$.

Answer 1

\begin{align}(1-a)^{n+1}&=(1-a)(1-a)^n\ge^{\text{true for $n$: (hypothesis)}}\\&\ge(1-a)(1-na)=\\&=1-na-a+na^2=\\&=1-(n+1)a+na^2\ge^{na^2\ge 0 \text{ for any value of $a$}}\\&\ge 1-(n+1)a\end{align}

Answer 2

At the end of this answer, it is proven by induction that $(1+x)^n\ge1+nx$ for $x\ge-1$. Thus, you are correct that your statement is true for $a\lt1$; that is, there is no need for $a\gt0$.

Answer 3

This is Bernoulli's inequality, and it is a special case of the general: $\forall n \in \mathbb N$

$$\prod_{i=1}^{n}(1-a_i)\ge 1-\sum_{i=1}^{n} a_i$$

for $0 \le a_i < 1$, where $\forall \ i \in \{1, 2, ..., n\}, a_i = a$

Read more:

https://cornellmath.wordpress.com/2008/01/26/convergence-of-infinite-products/

https://en.wikipedia.org/wiki/Infinite_product#Convergence_criteria

Show that $\prod (1- P(A_n))=0$ iff $\sum P(A_n) = \infty$

$\prod\limits_{k = 0}^\infty {(1 - {p_k})} = 0$ if and only if $\sum\limits_{k = 0}^\infty {{p_k}} $ diverges

Infinite product problem