Recurrence Inequality Define sequence ${a_n}$ as follows:
$a_1 = \frac {1}{2}, 2ka_k = (2k-3)a_{k-1}.$
Show that for any natural number $n, \sum_{k=1}^{n} a_k <1$ (No calculus is allowed).
I'm really just stuck on the right way to approach this. It is sufficient to show that $\sum_{k=1}^{\infty} a_k <1,$ but this isn't a series I am familiar with. I tried defining a partial sum series $b_n$ such that $b_n = \sum_{k=n}^{\infty} a_k,$ but that didn't help to find a non-trivial relationship.
 A: Define $b_k \equiv 2ka_k$, then for $k\geq 2$
$$b_k = (2k-3)a_{k-1} = 2(k-1)a_{k-1} - a_{k-1} = b_{k-1}-a_{k-1}$$
which rearranges to
$a_{k-1} = b_{k-1} - b_{k}$, so
$$
\sum_{k=2}^{n} a_{k-1}
= \sum_{k=2}^{n} (b_{k-1} - b_{k})
= b_1 - b_{n}
= 2(1)a_1 - 2na_n
= 1 - 2na_n$$
It is clear that $a_n \in (0,1)$ for all $n$, as $a_1=\tfrac12$ and inductively
$$a_n = \underbrace{\frac{2n-3}{2n}}_{\substack{\in (0,1)\\ \text{for $n\geq 2$}}}\underbrace{a_{n-1}}_{ \in(0,1)} \in (0,1)$$
Hence $$0<\sum_{k=1}^{n} a_k = 1 - 2\underbrace{(n+1)a_{n+1}}_{>0} < 1$$
A: Hint: $\;\sum a_k\;$ telescopes nicely:
$$
\begin{align}
a_k &= \frac{2k-3}{2k}a_{k-1} \\
 &= \frac{2k-3}{2k} \frac{2k-5}{2k-2} a_{k-2} \\
 &\ldots \\
 &= \frac{(2k-3)!!}{\frac{1}{2}(2k)!!} a_1 \\
 &= \frac{(2k-3)!!}{(2k)!!} \\
 &= \frac{\color{red}{\big(2k - (2k-1)\big)} \cdot (2k-3)!!}{(2k)!!} \\
 &= \frac{(2k-3)!!}{(2k-2)!!} - \frac{(2k-1)!!}{(2k)!!}
\end{align}
$$
A: Note that for $n\geq 1$,
$$A_n=\frac{1}{2}\prod_{k=2}^n\left(1-\frac{3}{2k}\right)=\frac{2\binom{2n-2}{n-1}}{n4^n}=\frac{2C_{n-1}}{4^n},$$
where $C_n$ is the $n$-th Catalan number.
Now 
$$\sum_{k=1}^n A_k=2\sum_{k=1}^n\frac{C_{k-1}}{4^k}=\frac{1}{2}\sum_{k=0}^{n-1}\frac{C_{k}}{4^k}=1-\frac{\binom{2n}{n}}{4^n}<1.$$
where the last equality can be easily proved  by induction.
P.S. The inequality has a combinatorial interpretation: Computing the sum of a Catalan sequence-- Random-walk motivated
