Proving the Existence of m for (1-1/m)^n > 1-ε in Real Analysis

[BrownianMan]

Let n ≥ 1 be an integer and ε > 0 a real number. Without making reference or use of nth roots, prove that there exists a positive integer m such that

$$\left (1- \frac{1}{m} \right )^{n}> 1-\varepsilon$$

How would I go about proving this? Would I just solve for m?

 

[tiny-tim]

Hi BrownianMan!

How would I go about proving this? Would I just solve for m?

Yup!

 

[SammyS]

Let n ≥ 1 be an integer and ε > 0 a real number. Without making reference or use of nth roots, prove that there exists a positive integer m such that
$$\left (1- \frac{1}{m} \right )^{n}> 1-\varepsilon$$
How would I go about proving this? Would I just solve for m?

Just solving for m would require use of an n^th root, right ? ... So, "No." to that.

 

[Ray Vickson]

Let n ≥ 1 be an integer and ε > 0 a real number. Without making reference or use of nth roots, prove that there exists a positive integer m such that

$$\left (1- \frac{1}{m} \right )^{n}> 1-\varepsilon$$

How would I go about proving this? Would I just solve for m?

What are you allowed to use? For example, using Calculus you can derive the inequality $$(1-x)^n > 1-nx ,$$ for $0 < x < 1.$

RGV

 

[BrownianMan]

Can't I say

$$\left ( 1-\frac{1}{m} \right )^{n} =1-\sum_{k=1}^{n}\binom{n}{k}\left ( \frac{1}{m} \right )^{k}=1-\frac{1}{m}\left (\sum_{k=1}^{n}\binom{n}{k} \left (\frac{1}{m} \right )^{k-1} \right )$$
$$\geq 1-\frac{1}{m}\left (2^{n}-1 \right )$$

So then

$$1-\frac{1}{m}\left (2^{n}-1 \right ) > 1 - \varepsilon$$
$$\frac{2^{n}-1}{\varepsilon }<m$$

 

[BrownianMan]

^Would this be right?

 

[tiny-tim]

Hi BrownianMan!

Can't I say

$$\left ( 1-\frac{1}{m} \right )^{n} =1-\sum_{k=1}^{n}\binom{n}{k}\left ( \frac{1}{m} \right )^{k}=1-\frac{1}{m}\left (\sum_{k=1}^{n}\binom{n}{k} \left (\frac{1}{m} \right )^{k-1} \right )$$
$$\geq 1-\frac{1}{m}\left (2^{n}-1 \right )$$

(Your ∑ needs a (-1)^k inside it. )

Perhaps I'm missing the obvious, but where does your 2ⁿ come from?