Prove (n)^(1/n) < 1 + sqrt(2/n) for all positive n.

[PiAreSquared]

Homework Statement

Prove:

(n)^(1/n) < 1 + sqrt(2/n) for all positive n.

Homework Equations

The Attempt at a Solution

Using induction, base case is easy enough to prove, however proving it holds for n+1 is where I am hitting a wall. So the problem is reduced to proving:

(n+1)^(1/(n+1)) < 1 + sqrt(2/(n+1)) for positive n.

 

[Tomer]

Hey,
it's pretty hard to deal with powers of 1/n, which is why you should first take the whole inequality in the power of n. Prove that for any n: n &lt; (1 + \sqrt{\frac{2}{n}})^n

Try to then work with it and see if you can make it. It takes a few steps, but it works. In the process I used another mini induction one has to prove (and can do so very easily).

If you prove this, then by taking the n'th root of this inequality and claiming that the arguments inside are all bigger the one, and therefore the inequality remains "<", you get the original one.

 

[BrownianMan]

I have a question based on this. Is it true that n^(1/n) - 1 <= sqrt(2/(n-1))? How do you show this?

And what is the mini induction you have to prove?

 

[jbunniii]

I have a question based on this. Is it true that n^(1/n) - 1 <= sqrt(2/(n-1))? How do you show this?

Your inequality is equivalent to
$$n \leq \left(1 + \sqrt{\frac{2}{n-1}}\right)^n$$
Try expanding the right hand side using the binomial theorem. Assuming $n \geq 2$, the first few terms are
$$\left(1 + \sqrt{\frac{2}{n-1}}\right)^n = 1 + \binom{n}{1} \sqrt{\frac{2}{n-1}} + \binom{n}{2} \frac{2}{n-1} + \cdots$$
What can you conclude?

 

[BrownianMan]

I though of another way. Induction. Base case is easy. To prove for n=k+1, we have

k+1 <= (1+ sqrt(2/k))^(k+1)

We have (1+ sqrt(2/k))^k*(1+ sqrt(2/k)) =(1+ sqrt(2/k))^(k+1)

Then I can say that (1+ sqrt(2/k))^k <=(1+ sqrt(2/k))^k*(1+ sqrt(2/k)) and since

k+1 <= (1+ sqrt(2/k))^k

by induction, the inequality holds.