MHB Problem involving discrete product

[anemone]

Hi members of the forum,

Please consider the following:

Given $\displaystyle a_{n}=\frac{n^2+1}{\sqrt{n^2+4}}$ where $\displaystyle n\in\mathbb{N}$

and $\displaystyle b_n=\prod_{k=1}^n(a_k)$

prove that $\displaystyle \frac{b_{n}}{\sqrt{2}}=\frac{\sqrt{n^2+1}}{\sqrt{n^2+2n+2}}$.

Therefore, deduce that
$\displaystyle \frac{1}{n^3+1}\le \frac{b_{n}}{\sqrt{2}}-\frac{n}{n+1} \le \frac{1}{n^3}$

I've been able to solve the first part of the question by 'observation' but not the second part of the problem. I just have no idea at all how to even start the deduction from the first part.

If anyone has any suggestions I'm all ear. Thanks.

BTW, here is my not-so-elegant proof to the first part of the question:

First, I have written out the product of the first few terms of $\displaystyle a_n$, to check if any simplification could be done to the expression and I get:$\displaystyle b_n=a_1 \times a_2 \times a_3 \times a_4 \times a_5 \times a_6 ...\times a_n$

$\displaystyle b_n=\frac{2}{\sqrt{5}}\times \frac{5}{\sqrt{20}} \times \frac{10}{\sqrt{85}} \times \frac{17}{\sqrt{260}} \times \frac{26}{\sqrt{626}} \times \frac{37}{\sqrt{1300}}\times ... \times a_n$

$\displaystyle b_n=\frac{1}{\sqrt{5}}\times \frac{1}{\sqrt{\frac{20}{2^2}}} \times \frac{1}{\sqrt{\frac{85}{5^2}}} \times \frac{1}{\sqrt{\frac{260}{{10}^2}}} \times \frac{1}{\sqrt{\frac{629}{{17}^2}}} \times \frac{1}{\sqrt{\frac{1300}{{26}^2}}} \times ... \times a_n$

$\displaystyle b_n=\frac{1}{\sqrt{5}}\times \frac{1}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{17}} \times \frac{\sqrt{5}}{\sqrt{13}} \times \frac{\sqrt{17}}{\sqrt{37}}\times \frac{\sqrt{13}}{\sqrt{5}}\times... \times a_n$$\displaystyle b_n=\frac{1}{\cancel {\sqrt{5}}}\times \frac{1}{\cancel {\sqrt{5}}} \times \frac{\cancel {\sqrt{5}}}{\cancel {\sqrt{17}}} \times \frac{\cancel {\sqrt{5}}}{\cancel {\sqrt{13}}} \times \frac{\cancel {\sqrt{17}}}{\cancel {\sqrt{37}}}\times \frac{\cancel {\sqrt{13}}}{\sqrt{5}}\times ... \times a_n$

So, if I want to find the product of the first 5 terms of $\displaystyle a_n$, I have

$\displaystyle b_5=a_1 \times a_2 \times a_3 \times a_4 \times a_5 $

$\displaystyle b_5=\frac{2}{\sqrt{5}}\times \frac{5}{\sqrt{20}} \times \frac{10}{\sqrt{85}} \times \frac{17}{\sqrt{260}} \times \frac{26}{\sqrt{626}}$

$\displaystyle b_5=\frac{1}{\sqrt {5}} \times \frac{1}{\sqrt {5}}\times \frac{\sqrt {5}}{\sqrt {17}}\times \frac{\sqrt {5}}{\sqrt {13}}\times \frac{\sqrt {17}}{\sqrt {37}}(26)$

$\displaystyle b_5= \frac{26}{\sqrt{13}.\sqrt{37}}=\frac{2\sqrt{26}}{\sqrt{37}}$.

$\displaystyle b_5=\sqrt{2}\left(\frac{\sqrt{numerator \; of \; a_5}}{\sqrt{numerator \; of \; a_6}}\right)$

Hence, I get $\displaystyle b_n=\sqrt{2}\left( \frac{\sqrt{numerator \; of \; a_n}}{\sqrt{numerator \; of \; a_{n+1}}}\right)=\sqrt{2}\left(\frac{\sqrt{n^2+1}}{\sqrt{(n+1)^2+1}}\right)=\sqrt{2}\frac{\sqrt{n^2+1}}{\sqrt{n^2+2n+2}}$

Or,

$\displaystyle \frac{b_n}{\sqrt{2}}=\frac{\sqrt{n^2+1}}{\sqrt{n^2+2n+2}}$

I hope I haven't confused you with my solution. If you have another way to prove this, could you please let me know too?

Thanks in advance.

 

[Deveno]

i believe you have an error in your post and you meant to define:

$a_n = \dfrac{n^2 + 1}{\sqrt{n^4 + 4}}$

with this in mind, here is an inductive proof of the first part of your problem, using induction on $n$:

clearly:

$b_1 = a_1 = \dfrac{2}{\sqrt{5}}$, so

$\dfrac{b_1}{\sqrt{2}} = \dfrac{\sqrt{2}}{\sqrt{5}} = \dfrac{\sqrt{1^2 + 1}}{\sqrt{1^2 + 2 + 2}}$

suppose that for $k$, we have:

$\dfrac{b_k}{\sqrt{2}} = \dfrac{\sqrt{k^2 + 1}}{\sqrt{k^2 + 2k + 2}}$

then:

$\displaystyle \frac{b_{k+1}}{\sqrt{2}} = \frac{1}{\sqrt{2}}\prod_{j=1}^{k+1} a_j = \frac{1}{\sqrt{2}}\left(\prod_{j=1}^k a_j\right)a_{k+1} = \frac{b_k}{\sqrt{2}}a_{k+1}$, and by our induction hypothesis:

$\displaystyle = \frac{\sqrt{k^2 + 1}}{\sqrt{k^2 + 2k + 2}}a_{k+1} = \frac{\sqrt{k^2 + 1}}{\sqrt{k^2 + 2k + 2}}\cdot \frac{(k+1)^2 + 1}{\sqrt{(k+1)^4 + 4}}$

$\displaystyle = \sqrt{\frac{k^2 + 1}{(k+1)^4 + 4}}\cdot \sqrt{k^2 + 2k + 2}$

now hold that thought...

to simplify the algebra, let $u = k+1$. then $u^2 - 2u + 2 = k^2 + 2k + 1 - 2k - 2 + 2 = k^2 + 1$, so our ugly square root on the left becomes:

$\displaystyle \sqrt{\frac{u^2 - 2u + 2}{u^4+4}} = \frac{1}{\sqrt{u^2 + 2u + 2}}= \frac{1}{\sqrt{(k+1)^2 + 2(k+1) + 2}}$

so that:

$\displaystyle \frac{b_{k+1}}{\sqrt{2}} = \frac{\sqrt{(k+1)^2 + 1}}{\sqrt{(k+1)^2 + 2(k+1) + 2}}$, and the result follows by the principle of induction.

i will try to post more later.

 

[anemone]

Hi Deveno,

You're right, I meant $\displaystyle a_n=\frac{n^2+1}{\sqrt{n^2+4}}$...I'm sorry about that mistype of the equation.

Thank you so much for the thorough proof by induction...it's awesome!

I'm very looking forward to read what you are going to post later...thanks in advance.

 

[Deveno]

are you sure you have the final inequality written properly?

i can prove:

$\displaystyle \frac{b_2}{\sqrt{2}} - \frac{n}{n+1} \leq \frac{1}{n^3}$

but the other side is giving me trouble, and for example when $n = 2$:

$\displaystyle \frac{b_2}{\sqrt{2}} = \frac{1}{\sqrt{2}}$

but:

$\displaystyle \frac{1}{\sqrt{2}} - \frac{2}{3} \sim 0.0404 < \frac{1}{9}$

get back to me on this.

 

[anemone]

Hi Deveno, I'm pretty sure I have got everything typed in correctly and you can also check the question out from its original source here: http://www.nizworld.com/wp-content/uploads/2010/12/NizWorldIrishMathematicalOlympiad1991.pdf (Page 2, Second problem)

But you're right, the inequality doesn't hold for n=1 either, something seems wrong with the stated inequality, I think.

If we ignore the 'wrong' side of the inequality, could you please show me the way you approached the other side of the inequality?

Thanks.

 

[Deveno]

what i was able to prove is this:

$\displaystyle \frac{1}{(n+1)^3} < \frac{b_n}{\sqrt{2}} - \frac{n}{n+1} < \frac{1}{n^3}$

note the left-hand side is considerably smaller than $\dfrac{1}{n^3 + 1}$.

the way i proceeded was considering the reciprocals, and using the fact that:

$\displaystyle \frac{1}{\frac{a}{b} - \frac{c}{d}} = \frac{bd}{ad - bc}$

so what i actually set out to prove was:

$\displaystyle n^3 < \frac{(n+1)\sqrt{n^2 + 2n + 2}}{(n+1)\sqrt{n^2 + 1} - n\sqrt{n^2 + 2n + 2}} < (n+1)^3$

then i just worked on the expression in the middle. rationalizing the denominator, we obtain:

$\displaystyle \frac{(n+1)^2\sqrt{(n^2+2n+2)(n^2+1)} + (n^2+n)(n^2+2n+2)}{2n+1}$

$\displaystyle = \frac{(n^2+2n+1)\left(\sqrt{n^4+2n^3+3n^2+2n+1} \right) + n^4 + 3n^3+4n^2+2n}{2n+1}$

so what we want to prove is:

$\displaystyle 2n^4 + n^3 < (n^2 + 2n + 1)\left(\sqrt{n^4 + 2n^3 + 3n^2 + 2n + 1} \right) + n^4 + 3n^3 + 4n^2 + 2n < 2n^4 + 7n^3 + 9n^2 + 5n + 1$

now clearly $n^2 = \sqrt{n^4} < \sqrt{n^4 + 2n^3 + 3n^2 + 2n + 1}$

so:

$(n^2 + 2n + 1)\sqrt{n^4 + 2n^3 + 3n^2 + 2n + 1} + n^4 + 3n^3 + 4n^2 + 2n >$

$n^4 + 2n^3 + 2n^2 + n^4 + 3n^3 + 4n^2 + 2n = 2n^4 + 5n^3 + 6n^2 + 2n > 2n^4 + n^3$.

that's one inequality down.

for the other, note that:

$n^4 + 2n^3 + 3n^2 + 2n + 1 < n^4 + 2n^3 + 5n^2 + 4n + 4 = (n^2 + n + 2)^2$

so:

$(n^2 + 2n + 1)\sqrt{n^4 + 2n^3 + 3n^2+2n+1}+n^4 + 3n^3 + 4n^2 + 2n <$

$(n^2 + 2n + 1)(n^2 + n + 2) + n^4 + 3n^3 + 4n^2 + 2n = n^4 + 3n^3 + 4n^2 + 3n + 1 + n^4 + 3n^3 + 4n^2 + 2n$

$= 2n^2 + 6n^3 + 8n^2 + 5n + 2 < 2n^2 + 7n^3 + 9n^2 + 5n + 1$

which finishes the proof.

*************************

why i think the original posted problem has a typo:

even if we use the "low estimate" of $\sqrt{n^4 + 2n^3 + 3n^2 + 2n + 1} \sim n^2$

we have $2n^4 + 5n^3 + 6n^2+2n > 2n^4 + n^3 + 2n + 1 = (2n + 1)(n^3 + 1)$

************************
although i have made every effort to check my work, the algebra involved here made my head hurt, and as a consequence may contain errors (also i might have messed up on the latex somewhere).

 

[anemone]

Thanks so much, Deveno! :D

I really understand the concept of it now, and I truly appreciate your time and effort to help me out. I am also extremely impressed by your willingness to type all these equations in LaTeX too.

Thus, I am deeply grateful to you, more grateful than I can fully express.

Thanks again.(Smile)

 

[Fernando Revilla]

BTW, here is my not-so-elegant proof to the first part of the question:

An easier and more elegant way is to prove by induction the equivalent equality:

$$\displaystyle\sum_{k=1}^n \left(\log(k^2+1)-\frac{1}{2}\log (k^4+4)\right)=\dfrac{1}{2}\left(\log 2+\log (n^2+1)-\log (n^2+2n+2)\right)$$

 

[anemone]

An easier and more elegant way is to prove by induction the equivalent equality:

$$\displaystyle\sum_{k=1}^n \left(\log(k^2+1)-\frac{1}{2}\log (k^4+4)\right)=\dfrac{1}{2}\left(\log 2+\log (n^2+1)-\log (n^2+2n+2)\right)$$

It's really nice of you to offer this suggestion to me and I surely do appreciate it. Thanks, Fernando Revilla!

I give this site many thumbs up!(Smile)