Proving Inequality for Natural Numbers n>2

[gop]

Homework Statement

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

[tex]\prod_{k=1}^{n}\frac{k^{2}+2}{k^{2}+1}<3[/tex]

and
[tex]\prod_{k=1}^{n}\frac{k^{2}+2}{k^{2}+1}<\frac{3n}{n+1}[/tex]

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

[tex]\prod_{k=2}^{n}k^{2}+2<\prod_{k=1}^{n}k^{2}+1[/tex]

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

[tex](\prod_{k=2}^{n}k^{2}+2)<(\prod_{k=1}^{n}k^{2}+1)\cdot\frac{n^{2}+2n+2}{n^{2}+2n+3}[/tex]

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.