Proving the Increasing Nature of the Sequence x_n=(1+1/n)^n

[MathematicalPhysicist]

i need to prove that the sequence is increasing, i.e, x_n+1>=x_n
which translates into:
[(n+2)/(n+1)]^(n+1)>=[(n+1)/n]^n

i tried through induction but got stumbled:
n=k
[(k+2)/(k+1)]^(k+1)>= [(k+1)/k]^k

n=k+1
[(k+3)/(k+2)]^(k+2)>=[(k+2)/(k+1)]^(k+1)
now i need to prove the last inequality, and got baffled.
any help is appreciated, perhaps induction isn't the right way?

p.s
i know this sequence converges to e, so spare me the trivial details about this sequence.
thanks in advance.

 

[SteveRives]

The derivative is > 0

 

[rachmaninoff]

For n > 0, its derivative is > 0. Therefore increasing.

So how do you show that
$$\ln \left(1+\frac{1}{x} \right) > \frac{1}{x+1}, \, x>0$$

 

[MathematicalPhysicist]

forgot to say that derivatives aren't allowed, or at least haven't yet been covered, this problem is from here:
http://www.maths.mq.edu.au/~wchen/lnfafolder/fa02-sl.pdf
page 14, the last problem.

 

[Dr Avalanchez]

i need to prove that the sequence is increasing, i.e, x_n+1>=x_n
which translates into:
[(n+2)/(n+1)]^(n+1)>=[(n+1)/n]^n

i tried through induction but got stumbled:
n=k
[(k+2)/(k+1)]^(k+1)>= [(k+1)/k]^k

n=k+1
[(k+3)/(k+2)]^(k+2)>=[(k+2)/(k+1)]^(k+1)
now i need to prove the last inequality, and got baffled.
any help is appreciated, perhaps induction isn't the right way?

p.s
i know this sequence converges to e, so spare me the trivial details about this sequence.
thanks in advance.

Try with induction by blocks (blocks of 2^k).

 

[VietDao29]

Or I think you can also expand the terms out, something like:
$$x_n = \left( 1 + \frac{1}{n} \right) ^ n$$
So:
$$x_n = 1 + n \times \frac{1}{n} + \frac{n (n - 1)}{2!} \times \frac{1}{n ^ 2} + \frac{n (n - 1) (n - 2)}{3!} \times \frac{1}{n ^ 3} + ... + \frac{n (n - 1) (n - 2) ... (n - (n - 1))}{n!} \times \frac{1}{n ^ n}$$
$$= 1 + 1 + \frac{1}{2!} \times \frac{n}{n} \times \frac{n - 1}{n} + \frac{1}{3!} \times \frac{n}{n} \times \frac{n - 1}{n} \times \frac{n - 2}{n} + ... + \frac{1}{n!} \times \frac{n}{n} \times \frac{n - 1}{n} \times \frac{n - 2}{n} \times ... \times \times \frac{n - (n - 1)}{n}$$
$$= 1 + 1 + \frac{1}{2!} \times \left( 1 - \frac{1}{n} \right) + \frac{1}{3!} \times \left( 1 - \frac{1}{n} \right) \times \left( 1 - \frac{2}{n} \right) + ... + \frac{1}{n!} \times \left(1 - \frac{1}{n} \right) \times \left(1 - \frac{2}{n} \right) \times ... \times \left(1 - \frac{n - 1}{n} \right)$$
You can do the same and come up with:
$$x_{n + 1} = 1 + 1 + \frac{1}{2!} \times \left( 1 - \frac{1}{n + 1} \right) + \frac{1}{3!} \times \left( 1 - \frac{1}{n + 1} \right) \times \left( 1 - \frac{2}{n + 1} \right) + ... + \frac{1}{(n + 1)!} \times \left(1 - \frac{1}{n + 1} \right) \times$$
$$\times \left(1 - \frac{2}{n + 1} \right) \times ... \times \left(1 - \frac{n}{n + 1} \right)$$
So can you say that x_{n + 1} > x_n?
Viet Dao,

 

[MathematicalPhysicist]

vietdao, from this expansion the inequality does follow.
but the question how did you arrive to it?
have you used Newton's binomial?

 

[rachmaninoff]

Binomial Theorem: (can be proved inductively)

$$(a+b)^n=\sum_{i=0}^n \frac{n!}{i! (n-i)!}a^ib^{n-i}$$

here a=1, b=1/n

 

[MathematicalPhysicist]

well thank you all, i should have thought of Newton's binomial.