1 Answer

Pawan Samdani · Answer 1 · 2018-11-09T06:53:36+0000

$a_n=\sqrt{((2n-1)!)/((2n+1)!)}$

Notice that

$((2n-1)!)/((2n+1)!)=((2n-1)!)/((2n+1)(2n)(2n-1)!)=\frac1{2n(2n+1)}$

So as
$n\to\infty$ you have
$a_n\to0$ . Clearly
a_n

must converge.

The second sequence requires a bit more work.

$\begin{cases}a_1=\sqrt2\\a_n=\sqrt{2a_(n-1)}&\text{for }n\ge2\end{cases}$

The monotone convergence theorem will help here; if we can show that the sequence is monotonic and bounded, then
a_n

will converge.

Monotonicity is often easier to establish IMO. You can do so by induction. When
n=2

, you have

$a_2=√(2a_1)=√(2\sqrt2)=2^(3/4)>2^(1/2)=a_1$

Assume
$a_k\ge a_(k-1)$ , i.e. that
$a_k=\sqrt{2a_(k-1)}\ge a_(k-1)$ . Then for
n=k+1

, you have

$a_(k+1)=√(2a_k)=\sqrt{2\sqrt{2a_(k-1)}\ge\sqrt{2a_(k-1)}=a_k$

which suggests that for all

, you have
$a_n\ge a_(n-1)$ , so the sequence is increasing monotonically.

Next, based on the fact that both
$a_1=\sqrt2=2^(1/2)$ and
a_2=2^(3/4)

, a reasonable guess for an upper bound may be 2. Let's convince ourselves that this is the case first by example, then by proof.

We have

$a_3=\sqrt{2*2^(3/4)}=\sqrt{2^(7/4)}=2^(7/8)$

$a_4=\sqrt{2*2^(7/8)}=\sqrt{2^(15/8)}=2^(15/16)$

and so on. We're getting an inkling that the explicit closed form for the sequence may be
a_n=2^((2^n-1)/2^n)

, but that's not what's asked for here. At any rate, it appears reasonable that the exponent will steadily approach 1. Let's prove this.

Clearly,
a_1=2^(1/2)<2