# Proving that a sequence is within certain bounds

Elysian

## Homework Statement

Define a1=1, and for every n>1, an+1 = an + $\frac{1}{an}$. 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 an+1 > an. I tried finding a limit, by saying if lim$_{n\rightarrow∞}$an = L then lim$_{n\rightarrow∞}$an+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

## The Attempt at a Solution

dirk_mec1
Is this the complete question?

Elysian
Yeah it is.

Homework Helper
Gold Member
The first step is to get some idea of how fast an grows. If you think of the recurrence relation as a PDE, Δa = an+1 - an = 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.

Homework Helper
Gold Member
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.

I took an and looked at an+12. Which gave a formula an2 + 2 + $\frac{1}{ an+12}$. I compared an2 to 2n, and said that an2 > 2n for n$\geq$3. Then by induction proved that this is the case for an+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.