1. Can you solve this?

Posted this in the Extracurricular Secondary Maths thread, only to realise it was Secondary Maths . Re-posting here:

Define X(n) & Y(n) in the following manner:

X(1) = Y(1) = 1, otherwise:

X(n)
= 2X(n-1) if n is Even
= 2X(n-1) - 1 if n is Odd

Y(n)
= 2Y(n-1) + 1 if n is Even
= 2Y(n-1) if n is Odd

Examples:

X(n) | 1 | 2 | 3 | 6 | 11 | 22 | 43 | 86 | ...
Y(n) | 1 | 3 | 6 | 13 | 26 | 53 | 106 | 213 | ...
n | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | ...

The problem:

Find Y(n)/X(n) as n -> infinity.

Good luck

2. Re: Can you solve this?

I think 5/2, will write proof in a sec.

$First, we observe that since: \frac{2Y_n}{2X_n+1}\leq \frac{Y_{n+1}}{X_{n+1}}\leq \frac{2Y_n+1}{2X_n}, and both sequences tend to \infty, it suffices to establish the result for one parity.\\ \\ To this end, we define: \bar{X}_n=X_{2n-1},\bar{Y}_n=Y_{2n-1}.\\ \\ The recurrences in the question then give\\ \bar{X}_{n+1}=4\bar{X}_n-1,\bar{Y}_{n+1}=4\bar{Y}_n+2.\\ \\ We can replace these sequences by the asymptotically equivalent sequences \\ \tilde{X}_n=\bar{X}_n-c_1,\tilde{Y}_n=\bar{Y}_n-c_2.\\ \\ By choosing c_1=1/3,c_2=-2/3, these transformed sequences \tilde{X}_n,\tilde{Y}_n both become geometric progressions with common ratio 4. Hence their ratio is constant.\\ \\ This implies that the sought limit is \\ \\ L=\frac{\tilde{Y}_1}{\tilde{X}_1}=\frac{\bar{Y}_1-c_2}{\bar{X}_1-c_1}=\frac{5}{2} .$

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