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:

Plot of first 1000 numbers and 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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  • 1
    $\begingroup$ It seems like you want to study the recurrence relation $a_{n+1}:=f(a_n)$ with $a_0:=x$ and $f(y):=y+\frac1y$. $\endgroup$
    – nejimban
    Jan 14 at 9:21
  • 1
    $\begingroup$ If @nejimban is right, note $a_{n+1}^2-a_n^2>2$. $\endgroup$
    – J.G.
    Jan 14 at 10:07
  • 1
    $\begingroup$ $a_{n+1}-a_n=\frac{a_{n+1}-a_n}{(n+1)-n}=\frac{1}{a_n}\,\, \Rightarrow \,\, \frac{d a_n}{dn}\approx \frac{1}{a_n}\,\,\Rightarrow\,\, a_n\sim\sqrt2\sqrt n$ $\endgroup$
    – Svyatoslav
    Jan 14 at 10:28
  • $\begingroup$ You may look here $\endgroup$
    – jjagmath
    Jan 14 at 18:14
  • $\begingroup$ Amazing thank you everyone for your help! They've put me on the right path now :) $\endgroup$
    – luke
    Jan 16 at 10:06

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