A question with the sequence $e_{n}=\left(1+\frac{1}{n}\right)^{n}$ [closed]

Prove that

$a$) the following sequence is increasing $$e_{n}=\left(1+\frac{1}{n}\right)^{n},\quad n\ge1;$$

$b$) the inequality below holds

$$e_{n} \leq3,\quad n\ge1.$$

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• a) also has a combinatorial proof, see also this for further arguments. – t.b. Sep 9 '12 at 22:37
• @ t.b.: thank you for the link! – user 1357113 Sep 10 '12 at 8:58
• @downvoter: what motivates you to downvote such a question? – user 1357113 Jan 30 '13 at 12:37

For the first part use the binomial theorem and show that each component is non-decreasing and some are increasing.

For the second part you can use the same binomial expansion term by term to show that $$\left(1+\frac1n\right)^n<1+1+\frac12+\dots\frac 1{r!}+\dots<1+1+\frac 12+\dots\frac 1 {2^r}+\dots$$ and sum the geometric progression.

• this could be a way! Thanks (+1) – user 1357113 Sep 9 '12 at 19:56

In order to prove that the given sequence is strictly increasing, we are to demonstrate $e_{n+1} > e_n$:

$\bigg(1+ \dfrac{1}{n+1}\bigg)^{n+1} > \bigg(1 + \dfrac{1}{n} \bigg)^n.$

Let's rewrite the inequality above as:

$\bigg( \dfrac{1 + \dfrac{1}{n+1}}{ 1 + \dfrac{1}{n}} \bigg)^n > \dfrac{1}{1 + \dfrac{1}{n+1}}.$

The right-hand side equals

$\dfrac{1}{1 + \dfrac{1}{n+1}} = \dfrac{n+1}{n+2} = 1 - \dfrac{1}{n+2}.$

Now, let's focus on the left-hand side:

$\bigg( \dfrac{1 + \dfrac{1}{n+1}}{ 1 + \dfrac{1}{n}} \bigg)^n = \bigg( \dfrac{(n+2)/(n+1)}{(n+1)/n} \bigg)^n = \bigg( \dfrac{n(n+2)}{(n+1)^2} \bigg)^n = \bigg( 1 - \dfrac{1}{(n+1)^2}\bigg)^n.$

By the Bernoulli's inequality, the following holds:

$\bigg( 1 - \dfrac{1}{(n+1)^2}\bigg)^n \geq 1 - \dfrac{n}{(n+1)^2}$

Now it's purely technical to show the desired inequality

$1 - \dfrac{n}{(n+1)^2} > 1 - \dfrac{1}{n+2},$

because

$\dfrac{n}{(n+1)^2} < \dfrac{1}{n+2}.$

• @Chris'ssister: In my Calculus course (at least) Bernoulli's inequality is an example of proof by induction. No calculus methods needed! – Jyrki Lahtonen Sep 9 '12 at 20:22
• @Jyrki Lahtonen: actually, you're right. – user 1357113 Sep 9 '12 at 20:30