Proving the Convergence of (1+1/n)^n Sequence | Help Needed!

[sutupidmath]

sequence question. Need help!

i would like to know where could i find the proof that the sequence

a_n=(1+1/n)^n bounded (upper bounded) by 4.

or in general that this sequence is a convergent one??

i know the proof by expanding it using binominal formula(Newton formula), but i am looking for another proof, by using some other helping sequence, and than to tell that this sequence a_n=(1+1/n)^n is smaller than every term of the helpin sequence?

any help would be appreciated.

 

[mathman]

Have you looked at the power series for e?

 

[sutupidmath]

Have you looked at the power series for e?

do u mean expressing e using taylor formula??

 

[Gib Z]

I Think that's what he meant, but here's a definition of e that will help :)

$$e=\lim_{n\to\infty} (1+\frac{1}{n})^n$$.
So as n goes to infinity, a_n goes to e, which is less than 4.

However, you need to know that a_n < 4 for ANY n. You know a_1=2.

So to make sure for any positive integer n, 2<a_n<4, we show that that a_n is an monotonically increasing function for positive n.

 

[sutupidmath]

I Think that's what he meant, but here's a definition of e that will help :)

$$e=\lim_{n\to\infty} (1+\frac{1}{n})^n$$.
So as n goes to infinity, a_n goes to e, which is less than 4.

However, you need to know that a_n < 4 for ANY n. You know a_1=2.

So to make sure for any positive integer n, 2<a_n<4, we show that that a_n is an monotonically increasing function for positive n.

Yes, i do know this. But what i am looking for is a proof(another proof, couse i already know two of them) using another sequence that will look something like this

b_n=(1+1/(n^2-1))^n

and than to show that for every n, a_n, is less than b_n, for every n.(a_n<b_n)

because i know that 2<e<3<4 .

because i need to prove it this whay(using another helping sequence which charasteristics we know).

thnx anywhay

 

[Gib Z]

Well I'm not sure about the form of your b_n, but here's another b_n that's larger term by term and is bounded by 4 as well.

$$e^x = \lim_{n\to\infty} (1+\frac{x}{n})^n$$
So a b_n that is larger term by term and bounded we be say..b_n = (1 + 1.002/n)^n? Any value of or less than ln 4 will do in place of 1.002.

 

[mathman]

e=1+1/1!+1/2!+1/3!+... which dominates the binomial expansion of (1+1/n)ⁿ

 

[sutupidmath]

e=1+1/1!+1/2!+1/3!+... which dominates the binomial expansion of (1+1/n)ⁿ

How does this help, to prove what i am looking for?