# Does a closed form expression of this series exist?

I have a function $$y=\frac{1}{x}$$. If $$x = 1$$, then $$y = 1$$; if $$x = 2$$, then $$y = 0.5$$, etc. If I wanted to add the previous term to the current term, I would have $$x = 1.5$$, $$x = 2.167$$ etc.

The terms in the sequence could be represented as:

Term 1:$$\ x+\frac{1}{x}$$

Term 2:$$\ x+\frac{1}{x+\frac{1}{x}}+\frac{1}{x}$$

Term 3:$$\ x+\frac{1}{x+\frac{1}{x}}+\frac{1}{x+\frac{1}{x+\frac{1}{x}}+\frac{1}{x}}+\frac{1}{x}$$

If the initial value was set to $$x = 1$$, the sequence would continue $$x = 2$$, $$x = 2.5$$, $$x = 2.9$$, with the recurrence relationship $$a_n =a_{n-1} +\frac{1}{a_{n-1} }$$ .

When I plot the first $$1000$$ numbers in the sequence, they converge to fit a power law line of best fit$$\ 1.4234x^{0.4993}$$, but I have no idea why. Shown below is the plot of the first $$1000$$ numbers and the line of best fit:

Does a closed form representation of this sequence exist? Something like a function of the Fibonacci numbers, or a sum of series, or a sum of series within a sum of series? Many thanks!

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• It seems like you want to study the recurrence relation $a_{n+1}:=f(a_n)$ with $a_0:=x$ and $f(y):=y+\frac1y$. Jan 14 at 9:21
• If @nejimban is right, note $a_{n+1}^2-a_n^2>2$.
– J.G.
Jan 14 at 10:07
• $a_{n+1}-a_n=\frac{a_{n+1}-a_n}{(n+1)-n}=\frac{1}{a_n}\,\, \Rightarrow \,\, \frac{d a_n}{dn}\approx \frac{1}{a_n}\,\,\Rightarrow\,\, a_n\sim\sqrt2\sqrt n$ Jan 14 at 10:28
• You may look here Jan 14 at 18:14
• Amazing thank you everyone for your help! They've put me on the right path now :)
– luke
Jan 16 at 10:06