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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.

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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