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Problem involving discrete product

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anemone

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Feb 14, 2012
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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.
 
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Deveno

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Feb 15, 2012
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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.
 
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anemone

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Feb 14, 2012
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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.:eek:

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

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Feb 15, 2012
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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.
 
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anemone

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Feb 14, 2012
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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

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MHB Math Scholar
Feb 15, 2012
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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).
 
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anemone

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Feb 14, 2012
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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

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MHB Math Helper
Jan 29, 2012
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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:

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

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Feb 14, 2012
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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)