If a sequence $\langle a_n\rangle$ is such that $a_1a_2=1, a_2a_3=2, \ldots$ and $\lim\frac{a_n}{a_{n+1}}=1.$ Then find $|a_1|.$ Here since $\lim \frac{a_n}{a_{n+1}}=1.$ So no definite conclusion can be made about the nature of the sequence $\langle a_n\rangle$.
So how can I can proceed to find the value of $a_1$ from the relation: $a_ka_{k+1}=k,$ for any $k\in\mathbb N$ ?
Please suggest something..
 A: You can easily get a uniqueness statement: suppose that $a_n$ and $b_n$ are two sequences satisfying the givens. Notice that $a_n$ and $b_n$ are never zero, because of the condition on products. Let $c_n = \frac{a_n}{b_n}$. Then we have $c_nc_{n+1} = 1$ and $\lim \frac{c_n}{c_{n+1}}=1$. If we let $c_1=x$, the first condition clearly specifies the whole sequence as
\begin{align*}
c_1=x && c_2=\frac{1}{x} && c_3 = x && c_4 = \frac{1}{x} && \ldots  
\end{align*}
The condition on the ratios is then that
\begin{align*}
x^2, \frac{1}{x^2}, x^2, \frac{1}{x}^2, \ldots \to 1 
\end{align*}
which gives $x= \pm 1$. So, either $a_n = b_n$ for all $n$, or else $a_n = -b_n$ for all $n$. Thus the solution, if it exists, is unique up to multiplying the whole sequence by $-1$ (which clearly preserves the given conditions).
A: Note that $a_{2k+1} = \frac{2k}{2k-1}a_{2k-1} = ... = a_1\prod_{j=1}^k \frac{2j}{2j-1} $  and $a_{2k+2} = a_2 \prod_{j=1}^k \frac{2j+1}{2j} = \frac{1}{a_1}\prod_{j=1}^k \frac{2j+1}{2j}$
By looking at the subsequence:
$$ 1 = \lim_{k \to \infty} \frac{a_{2k+1}}{a_{2k+2}} = a_1^2\lim_{k \to \infty} \frac{\prod_{j=1}^k \frac{2j}{2j-1}}{\prod_{j=1}^k \frac{2j+1}{2j}} = a_1^2 \lim_{k \to \infty} \prod_{j=1}^k \frac{(2j)(2j)}{(2j+1)(2j-1)} = a_1^2 \frac{\pi}{2} $$
Where we used Wallis Formula for $\pi$:
$$ \prod_{j=1}^\infty \frac{4j^2}{(2j-1)(2j+1)} = \frac{\pi}{2} $$
Similarly $$ 1 = \lim_{k \to \infty} \frac{a_{2k+2}}{a_{2k+3}} = \frac{1}{a_1^2} \lim_{k \to \infty} \frac{2k+1}{2k+2} \cdot\prod_{j=1}^k \frac{(2j+1)(2j-1)}{(2j)(2j)} = \frac{1}{a_1^2}\frac{2}{\pi}$$
Hence in both cases (and subsequences $(2k)_{k \in \mathbb N},(2k+1)_{k \in \mathbb N}$ cover whole sequence $(n)_{n \in \mathbb N}$), we get condition $$a_1^2 = \frac{2}{\pi}$$ so we get the value of $a_1$ up to the sign, that is: $$ |a_1| = \sqrt{\frac{2}{\pi}}$$
And Mike F already showed that $|a_1|$ is unique
