Solving $ a(n+1) = a(n) + \frac{1}{a(n)}$ with $a(1) = 1 $

Question

$ a(n+1) = a(n) + \frac {1}{a(n)}, a(1) = 1 $

What is the function that generates all the values of $a(n)$?

Upon first inspection, this function appears to lie somewhere between a fractional power of $k$ and a logarithmic function of $k$. My reason for suspecting logarithms is that the recurrence relation is harmonic. However, this is just a contemplation, and my other suspect possibility is a Taylor Series of some kind.

Edit: Echo provided me a link to OEIS, so now I will reveal the problem that generated this curiosity of mine.

Given $ a(n+1) = a(n) + \frac {1}{a(n)}, a(1) = 1 $

Prove $ a(100)>14 $

Answer 1

I don't think you can find an easy explicit formula for your recurrence relation. But there is an easy solution to you problem.

Consider the sequence $b_n = a_n^2$. Then we observe that $b_{n+1} = b_n + \frac{1}{b_n}+2>b_n+2$. So $b_n > 2n-1$ for all $n$. In particular, $$b_{100} > 199 > 196 = 14^2$$

Answer 2

I guess this idea is useful:

We know that $\sqrt {x+1} \simeq \sqrt {x} + \frac {1}{2 \sqrt {x}}$ and now if we set $f(x)= \sqrt{2x}$ then for large $x$ we have: $f(x+1) \simeq f(x)+\frac{1}{f(x)}$

For example: $a(100)=14.284...$ and $f(100)=\sqrt{200}=14.142...$
And it's not hard to show $\mid a(n)-f(n)\mid<1 \quad \forall n \in \mathbb N$ and for $n>2$ we have: $a(n)>f(n)$ (because $\sqrt {x} + \frac {1}{2 \sqrt {x}}> \sqrt {x+1})$