I'm required to show that the above series is decreasing. However, I encounter a problem when I realize that in the inductive step, I have a term for a(n) in both the numerator and denominator, which makes it difficult to show that a(n+1) > a(n+2). Any Help would be appreciated.
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Here’s a hint. Consider when the sequence would not decrease, namely where it would stay constant. To do this, solve:
$$x=\frac{x}{2}+\frac{1}{x}$$ $$x^2=2 \implies x=\{-\sqrt{2},\sqrt{2}\}$$
Since this sequence is always positive, the positive solution seems most relevant. As matter of fact, whenever $ a_n> \sqrt{2}$, $a_{n+1}$ decreases. I’ll leave it to you to rigorously prove the above claim, and show that $x>\sqrt{2} \implies \frac{x}{2}+\frac{1}{x}>\sqrt{2}$
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If $a_{n+1} =\dfrac{a_n}{2}+\dfrac1{a_n} $ then $a_{n+1}-a_n =-\dfrac{a_n}{2}+\dfrac1{a_n} =\dfrac{-a_n^2+2}{2a_n} $.
If $a_n > \sqrt{2}$ then $a_{n+1} < a_n$.
Also, $a_{n+1}^2 =\dfrac{a_n^2}{4}+1+\dfrac1{a_n^2} $ so $a_{n+1}^2-2 =\dfrac{a_n^2}{4}-1+\dfrac1{a_n^2} =(\dfrac{a_n}{2}-\dfrac1{a_n})^2 =(\dfrac{a_n^2-2}{2a_n})^2 $ so that $a_{n+1} > \sqrt{2}$.
Therefore if $a_n > \sqrt{2}$ then $\sqrt{2} < a_{m+1} < a_m $ for $m \ge n$.
Therefore, for any initial $a_1 > 0$, for all $n \ge 2$ we have $\sqrt{2} < a_{n+1} \lt a_n$ so $a_n \to^+ \sqrt{2}$.
You want to show that $a_{n+1}<a_n$ for all $n\geq1$. Writing this out yields $$\frac{a_n}{2}+\frac{1}{a_n}<a_n,$$ and subtracting $\frac{a_n}{2}$ from both sides leaves us with $$\frac{1}{a_n}<\frac{a_n}{2}.$$ This is equivalent to $a_n^2>2$. Can you prove this by induction?