Show that recursive sequence is decreasing 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) &gt; a(n+2). Any Help would be appreciated.
You want to show that $a_{n+1}&lt;a_n$ for all $n\geq1$. Writing this out yields
$$\frac{a_n}{2}+\frac{1}{a_n}&lt;a_n,$$
and subtracting $\frac{a_n}{2}$ from both sides leaves us with
$$\frac{1}{a_n}&lt;\frac{a_n}{2}.$$
This is equivalent to $a_n^2&gt;2$. Can you prove this by induction?
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&gt; \sqrt{2}$, $a_{n+1}$ decreases. I’ll leave it to you to rigorously prove the above claim, and show that $x&gt;\sqrt{2} \implies \frac{x}{2}+\frac{1}{x}&gt;\sqrt{2}$