# Let $n\geq 1$ and $\alpha \in [0,1)$ show that : $1\le \bigl(1+\frac{\alpha}{n}\bigr)^{n}\le \frac{1}{1-\alpha}$

Let $n\geq 1$ and $\alpha \in [0,1)$ show that : $$1\le \left(1+\dfrac{\alpha}{n}\right)^{n}\le \dfrac{1}{1-\alpha}$$

This question is related to that one Show that ${n \choose k}\leq n^k$

My thoughts:

To prove that the following statement, which we will call P(n), holds for all natural numbers n:

$$1\le \left(1+\dfrac{\alpha}{n}\right)^{n}\le \dfrac{1}{1-\alpha}$$

so my proof that P($n$) is true for each natural number $n$ proceeds as follows:

Basis:

Show that the statement holds for $n=1$.

P($1$) amounts to the statement:

$$1\le \left(1+\dfrac{\alpha}{1}\right)^{1}\le \dfrac{1}{1-\alpha}$$

$$\iff$$

$$1\le \left(1-{\alpha}^2\right)\le 1$$

since $\alpha \in [0,1) \implies 0\le \alpha < 1 \implies 0\le \alpha^2 < 1 \implies -1 \le -\alpha^2 < 0 \implies 0 \le 1-\alpha^2 < 1$ then the statement is true for $n=1$. Thus it has been shown that P($1$) holds

Inductive step:

Show that if P($n$) holds, then also P($n+1$) holds. This can be done as follows.

Assume P($n$) holds. It must then be shown that P($n+1$) holds, that is:

$$1\le \left(1+\dfrac{\alpha}{n+1}\right)^{n+1}\le \dfrac{1}{1-\alpha}$$

• I can't manage is my reasoning correct and is there other ways to prove that

Edit

since there is probleme in the case of P(1) becuase i shouldn't write $1\le \left(1-{\alpha}^2\right)\le 1.$ since the left inequality is not true. i have to break it in two separate case and do it then we ve :

• For $\left(1+\dfrac{\alpha}{n}\right)^{n}\le \dfrac{1}{1-\alpha}$

To prove that the following statement, which we will call P(n), holds for all natural numbers n:

$$\left(1+\dfrac{\alpha}{n}\right)^{n}\le \dfrac{1}{1-\alpha}$$

so my proof that P($n$) is true for each natural number $n$ proceeds as follows:

Basis:

Show that the statement holds for $n=1$.

P($1$) amounts to the statement:

$\left(1+\dfrac{\alpha}{1}\right)^{1}\le \dfrac{1}{1-\alpha}$ since $\alpha \in [0,1) \implies 0\le \alpha < 1 \implies 0\le \alpha^2 < 1 \implies -1 \le -\alpha^2 < 0 \implies 0 \le 1-\alpha^2 < 1$ then the statement is true for $n=1$. Thus it has been shown that P($1$) holds

Inductive step:

Show that if P($n$) holds, then also P($n+1$) holds. This can be done as follows.

Assume P($n$) holds. It must then be shown that P($n+1$) holds, that is:

$$\left(1+\dfrac{\alpha}{n+1}\right)^{n+1}\le \dfrac{1}{1-\alpha}$$

i can't manage

• For $1\le \left(1+\dfrac{\alpha}{n}\right)^{n}$
• It was wrong solution. $(1+x)^n\leq1+nx$ for $n\in(0,1)$. Commented Jan 4, 2015 at 16:29
• Isn't Bernoulli the opposite way? Commented Jan 4, 2015 at 16:31
• This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post.
– quid
Commented Jan 4, 2015 at 16:46

$$\left(1+\frac an\right)^n=\sum_{k=0}^n\binom nk\frac {a^k}{n^k}\le\sum_{k=0}^n a^k\le\sum_{k=0}^\infty a^k=\frac1{1-a}$$

Because e.g. $\binom nk=\frac{n(n-1)\ldots(n-k+1)}{k!}\le n^k$, but it's also possible to see this combinatorially. The last equality is the sum of a geometric series which works for $|a|<1$.

• yes i think this is way to prove it because is related to my last question. but let me see it what about the left one
– Educ
Commented Jan 4, 2015 at 16:26

Induction may not be the best approach here.

The left hand inequality is trivial. For the other, note than $\left(1-\frac\alpha n\right)^n\ge 1-\alpha$ by the Bernoulli inequality, hence $$\left(1+\frac\alpha n\right)^n\cdot(1-\alpha)\le \left(1+\frac\alpha n\right)^n\left(1-\frac\alpha n\right)^n =\left(1-\frac{\alpha^2} {n^2}\right)^n\le 1$$

By the Bernoulli inequality $\left\{\left(1+\frac{\alpha}{n}\right)^n\right\}_{n\geq 1}$ is an increasing sequence converging towards $e^\alpha$, so it is sufficient to prove that $e^{\alpha}\leq\frac{1}{1-\alpha}$, or: $$\int_{0}^{\alpha}1\,dz = \alpha \leq -\log(1-\alpha)=\int_{0}^{\alpha}\frac{dz}{1-z}.$$