Limit of a Recursive Expression

Calculus Level 2

True or False?

For any real k > 0 , k >0, given the recursive expression
x n = x n − 1 + k x n − 1 + 1 , x_{n} = \frac{x_{n-1}+k}{x_{n-1}+1}, if x 0 = 0 , x_0 = 0, then lim ⁡ n → ∞ x n = k . \displaystyle \lim_{n\to\infty} x_n= \sqrt{k}.

False True

Timothy Cao by Timothy Cao , 21, Canada

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1 solution

Timothy Cao
Feb 24, 2018

An amateur's quick explanation (just for ideas basically)

k = k + k k + 1 \sqrt{k} = \frac{\sqrt{k}+k}{\sqrt{k}+1}

k ( k + 1 ) = k + k \sqrt{k} (\sqrt{k}+1)= \sqrt{k}+k

k + k = k + k k+\sqrt{k} = k+\sqrt{k}

We can substitute x n x_n and x n − 1 x_{n-1} both with k k because if this series converges, then as n n approaches ∞ \infty , x n = x n − 1 = x_n = x_{n-1} = convergence value

We know this converges because the function f ( x ) = x + k x + 1 f(x) = \frac{x+k}{x+1} has only one positive value a a where ( a , a ) (a,a) is a coordinate on the graph.

*And that value is k \sqrt{k} *

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Forgive me if this is a dumb question, as I'm learning analysis for the first time. Aren't you assuming that { x n } \{ x_n \} converges for this behaviour to happen? For, if the sequence is unbounded above (it's non trivial that this sequence is bounded, imo), then you can't use this trick. So, I think one necessary step before you do this is prove that the sequence is bounded above and monotonically increasing (or if you have another way to show that the sequence does converge, do that).

Hobart Pao - 3 years, 1 month ago

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