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here is solution of my old question but i can't see it would someone explain to me the principal idea and what he wants to show

Solution

from $u_n=\sqrt{n}\prod_{k=1}^{n}\left(1-\frac{1}{2k}\right),$ To $u^2_n=\frac{n}{4}\prod_{k=2}^{n}\left(1-\frac{1}{k}+\frac{1}{4k^2}\right),$

and from $u^2_n= \frac{n}{4}\prod_{k=2}^{n}\left(1-\frac{1}{k}+\frac{1}{4k^2}\right)$ To $u^2_n=\frac{n}{4}\prod_{k=2}^{n}\left(1-\frac{1}{k}\right)\prod_{k=2}^{n}\left(1+\frac{1}{4k(k-1)}\right)$

and if we've that $u_n^2$ convergent is that means $u_n$ also convergent

i just need to see the principal idea of what he did Thanks

I think the principal idea of solution is that he want to show $u_n^2$ convergent and bounded then tell that $u_n$ is convergent too Any help would be appreciated

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Educ
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  • If you tell us what principale is and what you have tried yourself, maybe someone can help. – Karl Dec 24 '14 at 08:47
  • I think the principal idea of solution is that he want to show $u_n^2$ convergent and bounded then tell that $u_n$ is convergent too – Educ Dec 24 '14 at 08:52

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Note first that $$ 1-\frac{1}{k}+\frac{1}{4k^2} =\left(1-\frac{1}{k}\right)\left(1+\frac{k}{k-1}\cdot\frac{1}{4k^2}\right)= \left(1-\frac{1}{k}\right)\left(1+\frac{1}{4(k-1)k}\right). $$ Next, $$ \frac{n}{4}\prod_{k=2}^n\left(1-\frac{1}{k}\right)=\frac{n}{4}\prod_{k=2}^n \frac{k-1}{k}=\frac{1}{4}. $$ Then $$ \prod_{k=2}^n\left(1+\frac{1}{4k(k-1)}\right)=\exp\left(\sum_{k=2}^n \log\left(1+\frac{1}{4k(k-1)}\right)\right)\le\exp\left(\sum_{k=2}^n \frac{1}{4k(k-1)}\right), $$ and $$ \sum_{k=2}^n \frac{1}{k(k-1)}=\sum_{k=2}^n\left(\frac{1}{k-1}-\frac{1}{k}\right)=1-\frac{1}{n}<1. $$ Thus $$ \prod_{k=2}^n\left(1+\frac{1}{4k(k-1)}\right)\le \exp\left(\frac{1}{4}\right) $$ and hence $\{u_n^2\}$ is upper bounded and increasing, and thus convergent, and so is $\{u_n\}$, as $u_n>0$.

Yiorgos S. Smyrlis
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