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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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|cite|improve this question

closed as off-topic by kamil09875, Antonios-Alexandros Robotis, Jonas, Jean-Claude Arbaut, Ben S. Mar 22 at 6:02

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – kamil09875, Antonios-Alexandros Robotis, Jonas, Jean-Claude Arbaut, Ben S.
If this question can be reworded to fit the rules in the help center, please edit the question.

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 1618033 Sep 10 '12 at 8:58
@downvoter: what motivates you to downvote such a question? – user 1618033 Jan 30 '13 at 12:37
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|cite|improve this answer
this could be a way! Thanks (+1) – user 1618033 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

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


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

share|cite|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. – user 1618033 Sep 9 '12 at 20:30

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