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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.

dcxt
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  • Hint : $$\sum\limits_{k=1}^n \dfrac{1}{2^k} \lt 1$$

    and

    for $k\ge 2$ $$a_k = \dfrac{2k-3}{2k}a_{k-1} \lt \dfrac{1}{2k}a_{k-1} \lt \dfrac{1}{2}a_{k-1}$$

     – AgentS Jun 27 '18 at 18:16
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    @rsadhvika: $\dfrac{2k-3}{2k}a_{k-1} \lt \dfrac{1}{2k}a_{k-1}$ isn't true. – joriki Jun 27 '18 at 18:18
  • This seems like a rather calculus-y problem; it's a bit harsh not to allow calculus. Could you provide some context please? Who doesn't want you to use calculus? – joriki Jun 27 '18 at 18:20
  • Oh right, obviously it isn't true, I'll try again, ty :) – AgentS Jun 27 '18 at 18:20
  • This is a olympiad math problem, so calculus is technically allowed, but generally most competitors tend to stick to traditional algebra and inequalities because graders are especially harsh if a calculus solution is not perfect. – dcxt Jun 27 '18 at 18:29

3 Answers3

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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 $$0<\sum_{k=1}^{n} a_k = 1 - 2\underbrace{(n+1)a_{n+1}}_{>0} < 1$$

adfriedman
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Hint: $\;\sum a_k\;$ telescopes nicely:

$$ \begin{align} a_k &= \frac{2k-3}{2k}a_{k-1} \\ &= \frac{2k-3}{2k} \frac{2k-5}{2k-2} a_{k-2} \\ &\ldots \\ &= \frac{(2k-3)!!}{\frac{1}{2}(2k)!!} a_1 \\ &= \frac{(2k-3)!!}{(2k)!!} \\ &= \frac{\color{red}{\big(2k - (2k-1)\big)} \cdot (2k-3)!!}{(2k)!!} \\ &= \frac{(2k-3)!!}{(2k-2)!!} - \frac{(2k-1)!!}{(2k)!!} \end{align} $$

dxiv
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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

Robert Z
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