- #1

- 162

- 0

## Homework Statement

Show that [tex] lim_{n\rightarrow +\infty} { (1 + 1/(2n))^n } = \sqrt{e}[/tex].

## Homework Equations

I am allowed to assume * [tex] lim_{n\rightarrow +\infty} { (1 + 1/n )^n} = e[/tex].

I am

*not*allowed to use the theorem that asserts [tex]lim_{n\rightarrow n_0} {\sqrt{S_n}} = (lim_{n\rightarrow n_0} {S_n})^{1/2} [/tex].

## The Attempt at a Solution

I want to show that the sequence is increasing and bounded, and therefore the [tex]lim_{n\rightarrow +\infty} { (1 + 1/(2n))^n }[/tex] exists. Let's suppose I have show the sequence is increasing by comparing [tex] S_k[/tex] and [tex] S_{k+1} [/tex] and showing [tex] \forall k, S_k < S_{k+1}[/tex]. Let's also suppose I know that 2 is an upper bound for [tex] S_n[/tex]. Then I want to show [tex] \forall \epsilon >0[/tex] [tex] \exists N[/tex] such that [tex] \forall n > N[/tex], [tex]| (1 + 1/(2n) )^n - \sqrt{e} | < \epsilon[/tex].

Does this argument work?

*Lemma*: [tex] \forall \epsilon > 0[/tex] [tex]\exists N_0[/tex] such that [tex] \forall n > N_0 [/tex], [tex] |2 + 1/n - 2| < \epsilon [/tex]. That is, [tex] lim_{n\rightarrow +\infty} {(2 + 1/n)} = 2[/tex].

[tex] \forall n | ( 1 + 1/(2n) )^n - \sqrt{e} | < |1/n|[/tex]. Since [tex] \forall n, 1/n > 0[/tex], then [tex] |1/n| = 1/n[/tex]. By our

*lemma*, we know [tex] |1/n| = 1/n < \epsilon[/tex]. Choose an [tex] n > 1/ \epsilon [/tex]. Then [tex] | ( 1 + 1/(2n) )^n - \sqrt{e} | < \epsilon [/tex] and therefore [tex] lim_{n\rightarrow +\infty} { (1 + 1/(2n))^n } = \sqrt{e}[/tex].

I was originally trying to use * with the definition of a limit to show [tex] | ( 1 + 1/(2n) )^n - \sqrt{e} | < | ( 1 + 1/(2n) )^n - e | < \epsilon [/tex] but I couldn't figure out how to determine

*N*.