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From the literature, I have found the following: Any real number A (say) can be expressed as

$ A = a_1 + (1/a_1) + (1/a_2) + (1/a_3) +\ldots$ Where $a_1\ge2$ and the recurrence relation $a_{i+1}\ge a_i(a_i - 1) + 1$ for $i \ge 1$.

I could not understand this statement due to the fowling reason:

I fixed $a_1$ = 2 then, $A = 2 + ½ + (1/a_2) + (1/a_3) + \dots$ ---------(i)

We can find $a_2$ by recurrence relation; $a_2 \ge a_1(a_1 -1) + 1 = 2(2-1) + 1 = 3$

i.e., $a_2\ge 3$ and $a_3\ge4$ and so on… Now, by our (i), $A = 2 + 1/2 + 1/3 + 1/4 + \dots$

How any real number A is equal to (i)?

I could not understand this statement.

Also, the same A can be expressible in other two ways:

$A = a_0 + (1/a_1) + (1/a_1)(1/a_2) + (1/a_1)(1/a_2)(1/a_3)+ \ldots$ Where $a_1\ge2$ and recurrence relation $a_{i+1}\ge a_i$ for $i \ge1$.

Also, $A = a_0 + (1/a_1) + (1/(a_1-1)) (1/a_1) (1/a_2) \\+ (1/(a_1-1)) ((1/a_2-1)) (1/a_1)(1/a_2)(1/a_3) +\ldots$

Briefly, explain where I am wrong in my example or not? Also, how to deduce one equation to other equations of A? A waiting your explanations and proofs.

Edit Is it applicable for any GIVEN real number, instead of any real number? If yes, how to proceed to complete the proof? can we deduce from one representation to other representations?

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    @did Indeed not, and I was rather hoping that the person who asked the question would put in a little work to do this, rather than being told. There are some interesting things to investigate around these egyptian fraction type decompositions.2012-08-28

1 Answers 1

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I suspect that you are saying that real $A$ has a single representation as $A = a_0 + (1/a_1) + (1/a_2) + (1/a_3) +\cdots$ for integer $a_i$ where $a_1\ge2$ and the recurrence relation $a_{i+1}\ge a_i(a_i - 1) + 1$ for $i \ge 1$.

This is (at least for irrational $A$) often described as the Egyptian fraction representation: for example OEIS A001466 gives $\pi = 3 + \frac{1}{8} + \frac{1}{61} + \frac{1}{5020} + \frac{1}{128541455}+\cdots .$

To show it exists and is unique, consider the partial sum $A_i$ up to the $(1/a_i)$ term, where $a_0 = \lceil{A}\rceil - 1$ and $a_{i+1}=\left\lfloor{\frac{1}{A-A_i}}\right\rfloor +1 $.

A proof will use the following points:

  • $A_i \lt A \le A_{i-1} + \frac{1}{a_i - 1}$
  • $\frac{1}{n-1}-\frac{1}{n} = \frac{1}{n(n-1)}$
  • if $b_{j+1}= b_j(b_j - 1) + 1$ then $\sum_{j=k}^{\infty} \frac{1}{b_j} = \frac{1}{b_k -1}$

Your second expression can be handled in a similar way with suitable adjustments: for example the third bullet point becomes $\sum_{j=1}^{\infty} \left(\frac{1}{b_k}\right)^j = \frac{1}{b_k -1}$.

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    Henry Sir, can you complete any one of the proof of A?please...2012-08-29