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I need help using induction on a recursive sequence.

Given $S_{1}=2$ and $S_{n+1}= \frac{S_{n}}{2} + \frac{1}{S_{n}}$

I am working on the recursive convergence to $\sqrt{2}$, therefore I want to show that it is bounded below by an arbitrary lower bound, in which I chose 1. thus by induction I want to show that $S_{n+1} > 1$, $$ S_{n} > 1$$ $$\frac{1}{S_{n}} < 1$$ $$S_n +\frac{1}{S_{n}} > ?+1$$

I get stuck here. Im not to sure how to get to my end point of $S_{n+1}$.

Ever Olivares
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  • Why not bound it below by $0$？ – WhatsUp Oct 05 '19 at 02:52
  • It should not matter as much. As long as its a lower bound less than $\sqrt{2}$ the statement should hold for any lower bound. nonetheless if I did use $0$, I wouldn't know what to do in the third step. – Ever Olivares Oct 05 '19 at 02:56
  • That was my point: if you use $0$, then the third step is trivial ($S_n > 0$ implies $\frac{S_n}{2} + \frac{1}{S_n} > 0$)! Hence it's done. – WhatsUp Oct 05 '19 at 02:59
  • by induction to get there you'd have to algebraically manipulate the inequalities, to indeed show that is true. The trouble I'm having here is that there is two terms of $S_n$ making the algebra not so straight forward – Ever Olivares Oct 05 '19 at 03:06
  • Hint: $AM/GM{}{}{}{}{}$ – Thomas Andrews Oct 05 '19 at 03:09
  • I have seen in other posts mention AM/GM, the professor has never proved it or even mentioned it. I've actually never seen it until today. Therefore, would there be another way to approach it, and if not would you be able to explain the AM/GM is applied to my inequality. – Ever Olivares Oct 05 '19 at 03:17

2 Answers2

1

As for all $n$, $S_n>0$,

By AM GM inequality,

$S_{n+1}=\frac{S_n}{2}+\frac{1}{S_n}\ge2\sqrt{\frac{S_n}{2}\bullet\frac{1}{S_n}}=\sqrt2$

So $S_n\ge\sqrt2>1.$

Culver Kwan
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  • Regarding my real analysis class currently. AM GM is not in my arsenal. However thank you for providing a thorough response on how its applied to my inequality! – Ever Olivares Oct 05 '19 at 03:27
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    AMGM inequality for two values follows from $ (a-b)^2\geq0$ because then $a^2+b^2\geq 2ab$. With $a=\sqrt{x}, b=\sqrt{y}$ you get the desired inequality. – Jens Schwaiger Oct 05 '19 at 03:50
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$S_n>1$

$S_n-1>0$

$(S_n-1)^2>0$

$S_n^2-2S_n+1>0$

$S_n^2+1>2S_n$

$\frac{S_n}{2}+\frac{1}{2S_n}>1$

$S_n<2S_n$

$\frac{1}{S_n}>\frac{1}{2S_n}$

$\frac{S_n}{2}+\frac{1}{S_n}>\frac{S_n}{2}+\frac{1}{2S_n}>1$

quantus14
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