Proving Inequality for Natural Numbers n>2

gop
Messages
55
Reaction score
0

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...
 
Physics news on Phys.org
You could try writing the product as the exponential of a sum, and then bounding the sum by an integral.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top