Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

share|improve this question
1  
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! –  Chris's sis Sep 10 '12 at 8:58
    
@downvoter: what motivates you to downvote such a question? –  Chris's sis Jan 30 '13 at 12:37
add comment

2 Answers 2

up vote 4 down vote accepted

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.

share|improve this answer
    
this could be a way! Thanks (+1) –  Chris's sis Sep 9 '12 at 19:56
add comment

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

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

share|improve this answer
    
@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. –  Chris's sis Sep 9 '12 at 20:30
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.