Need guidance on my proof of limit.

[Samuelb88]

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.

 

[Office_Shredder]

It's not clear to me what the relevance of the lemma is to picking your value of n (why can't you just pick n larger than one over epsilon without looking at a different limit?), or why [tex]
\forall n | ( 1 + 1/(2n) )^n - \sqrt{e} | < |1/n|
[/tex] is true.

[tex]
| ( 1 + 1/(2n) )^n - \sqrt{e} | < | ( 1 + 1/(2n) )^n - e | < \epsilon
[/tex]

This middle term in the inequality is going to fairly large (in the analysis sense) when n is large. We know [tex]( 1 + 1/(2n) )^n[/tex] is going to be close to [tex] \sqrt{e}[/tex], so it can't be close to e as well. However, at some point you are going to need to use the limit that gives e, or some alternative definition of e

 

[Samuelb88]

It's not clear to me what the relevance of the lemma is to picking your value of n (why can't you just pick n larger than one over epsilon without looking at a different limit?), or why [tex] \forall n | ( 1 + 1/(2n) )^n - \sqrt{e} | < |1/n|[/tex] is true.

I added the lemma to use the fact I knew the limit of the 2+1/n was 2, and to bound the difference [tex]|(1 + 1/(2n))^n - \sqrt{e}| < |1/n|[/tex] for all n (n is of course restricted to naturals, and this doesn't include zero). So my reasoning was after establishing the above inequality, to use the lemma to argue I can find an n>N such that [tex]|1/n| = 1/n < \epsilon[/tex]. That such n being any integer [tex] > 1/\epsilon[/tex]. Therefore if [tex] n > 1/\epsilon[/tex], then [tex]|(1 + 1/(2n))^n - \sqrt{e}| < \epsilon[/tex]. Does this make sense?

This middle term in the inequality is going to fairly large (in the analysis sense) when n is large. We know [tex]( 1 + 1/(2n) )^n [/tex] is going to be close to [tex]\sqrt{e}[/tex] , so it can't be close to e as well. However, at some point you are going to need to use the limit that gives e, or some alternative definition of e

Sorry, I meant to say: [tex] |(1 + 1/(2n)^n - \sqrt{e}|[/tex] < [tex]|(1+1/n)^n - e|[/tex]. Then using the fact I that I knew [tex] \forall \epsilon[/tex] there was some N such that whenever n>N, I was trying to argue [tex]|(1+1/n)^n - e| < \epsilon[/tex]. The main problem I had was deriving an explicit N from the inequality above. That is why I choose to bound the difference [tex] |(1 + 1/(2n)^n - \sqrt{e}|[/tex] by [tex]1/n[/tex] instead.

 

[Samuelb88]

Here is an idea that I just thought of using the fact that I may assume * [tex] \lim_{n/rightarrow +/infty} {(1+1/n)^n}=e[/tex]. I am not sure if it is considered "rigorous" enough. Here it is:

I may use the fact that ** if [tex]S_n, T_n[/tex] are sequences, and [tex] \lim_{n\rightarrow +\infty} {S_n} =A[/tex], [tex] \lim_{n\rightarrow} {T_n} = B[/tex]. Then [tex] \lim_{n\rightarrow} {S_n T_n} = AB[/tex].

Suppose * and let m=2n. Then [tex] \lim_{n\rightarrow +\infty} {(1 + 1/n)^n} = \lim_{m\rightarrow +\infty} {(1+1/(2m))^{2m}[/tex] (Here I am making the argument that as [tex] n\rightarrow +\infty[/tex], [tex] 2m\rightarrow +\infty[/tex] and hence [tex] m\rightarrow +\infty[/tex]). By **, [tex] \lim_{n\rightarrow +\infty} {(1+1/(2m)^m (1+1/(2m))^m } = \lim_{m\rightarrow +\infty} {(1+1/(2m))^m} \lim_{m\rightarrow +\infty} {(1+1/(2m))^m = e[/tex]. Then [tex] \lim_{n\rightarrow +\infty} {(1+1/(2m))^m } = \sqrt{e} [/tex].