Proving Inequality for Natural Numbers n>2

[gop]

Homework Statement

Proof that for n>2 and n is a natural number it holds that

$$\prod_{k=1}^{n}\frac{k^{2}+2}{k^{2}+1}<3$$

and
$$\prod_{k=1}^{n}\frac{k^{2}+2}{k^{2}+1}<\frac{3n}{n+1}$$

Homework Equations

The Attempt at a Solution

My best approach was to split the product over the fraction and then to arrive at a statement that looks like

$$\prod_{k=2}^{n}k^{2}+2<\prod_{k=1}^{n}k^{2}+1$$

I then tried to prove by induction that this statement holds but that doesn't really work. The best result I got (for n+1) is then

$$(\prod_{k=2}^{n}k^{2}+2)<(\prod_{k=1}^{n}k^{2}+1)\cdot\frac{n^{2}+2n+2}{n^{2}+2n+3}$$

But I can't do anything usefuel with that...

 

[Avodyne]

You could try writing the product as the exponential of a sum, and then bounding the sum by an integral.