Proving that a sequence is within certain bounds

[Elysian]

Homework Statement

Define a₁=1, and for every n>1, a_n+1 = a_n + [itex]\frac{1}{a_n}[/itex]. Prove that 20 < a₂₀₀ < 24

The Attempt at a Solution

I tried a few things to no avail. First, I showed that this is an increasing function by showing a_n+1 > a_n. I tried finding a limit, by saying if lim$_{n\rightarrow∞}$a_n = L then lim$_{n\rightarrow∞}$a_n+1 = L. Plugging in L, and solving but that didn't help. I feel like I need to find an upper and lower bound and work from there but I'm not sure how

Homework Statement

Homework Equations

The Attempt at a Solution

 

[dirk_mec1]

Is this the complete question?

 

[Elysian]

Yeah it is.

 

[haruspex]

The first step is to get some idea of how fast a_n grows. If you think of the recurrence relation as a PDE, Δa = a_n+1 - a_n = 1/a then you can integrate to get a as a function of n. You'll then see how the n=200 relates to the lower bound of 20. Next, need to make up some inductive hypothesis for how far the sequence can deviate from the analytic solution.

 

[haruspex]

Hi Elysian... Have you solved this now, got too busy elsewhere, or given up? Did you understand my hint? I can guide you to the solution.

 

[Elysian]

Hi Elysian... Have you solved this now, got too busy elsewhere, or given up? Did you understand my hint? I can guide you to the solution.

I did actually solve it, thank you for checking back. My bad on not replying. I found an upper and lower bound of the sequence, and with this I plugged in 200 into the lower bound of sqrt(2n), and the upper bound of sqrt(13n/6) and then got 20 and 20.866 respectively.

 

[CompuChip]

Of course 20.866 is certainly smaller than 24, but it seems a bit tight given the number in the question. This makes me a bit suspicious. Are you sure about your calculation? :-)

 

[Elysian]

Of course 20.866 is certainly smaller than 24, but it seems a bit tight given the number in the question. This makes me a bit suspicious. Are you sure about your calculation? :-)

Pretty sure. To give you an overview of what I did,

I took a_n and looked at a_n+1². Which gave a formula a_n² + 2 + [itex]\frac{1}{ a_n+1²}[/itex]. I compared a_n² to 2n, and said that a_n² > 2n for n$\geq$3. Then by induction proved that this is the case for a_n+1² $\geq$ 2(n+1). Then I had to find a value greater than 2n, so I approximated that it would be (2+z)n, where z is a small number. Finding z to be 1/6, I plugged that in, got 13/6 as 2+z, and then continued the same process as before.