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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  • 1
    $\begingroup$ a) also has a combinatorial proof, see also this for further arguments. $\endgroup$ – t.b. Sep 9 '12 at 22:37
  • $\begingroup$ @ t.b.: thank you for the link! $\endgroup$ – user 1357113 Sep 10 '12 at 8:58
  • $\begingroup$ @downvoter: what motivates you to downvote such a question? $\endgroup$ – 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.

  • $\begingroup$ this could be a way! Thanks (+1) $\endgroup$ – 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

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


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

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

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