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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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    Could you tell us where "in the literature" you found this? It might help us if we could see where this comes from.2012-08-28
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    The point is not that every series of the form you have given represents a real number, but that any real number can be expressed in the series form. As you have shown there are possible series which don't express real numbers.2012-08-28
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    The first series is unusual and your later examples have different notation - should it begin with $a_0$?2012-08-28
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    If we are to take this at face value, real numbers $\le2$ don't exist. Curious.2012-08-28
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    On a side note, your use of TeX is really odd. You're supposed to put dollar signs around entire formulas, not just around small bits of formula. And you can get $\ge$ by typing `\ge`. I fixed it up as best I could; please look it over to make sure I did not accidentally change the meaning.2012-08-28
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    OLD JOHN SIR, Literature means, I studied in my text as one of the note at the end of the chapter. That chapter name is writing real numbers in infinite series.2012-08-28
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    Yes, but which **book** does it come from? Who is the author?2012-08-28
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    HARALD HANCHE OLSEN SIR, I will use latex properly in next time. Thank you for editing the post.2012-08-28
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    MARK BENNET SIR, may be $a_0$ is need. I am not sure. but, can we express the same for any given real number?2012-08-28
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    As you have it there is no way of expressing zero (or any negative number, or indeed any real number less than 2.5) ... if you had $a_0$ the situation would be rather different, so you see it does matter what the statement is, and no-one will be able to answer unless you clarify the various matters noted in the comments.2012-08-28
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    @MarkBennet As a matter of fact, it is not so difficult to guess what would be a correct formulation of the first statement (and to prove it)... But I fully agree that the OP should first clarify (and maybe, o foolish hope, answer OldJohn's question about the source).2012-08-28
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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

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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, you have added some good flavor to my post. Can you prove the my initial statement? I confused. please do not give hints. Prove the fist statement and then I will try for other two statements of A.2012-08-28
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    Sir! can you make proof of (i) by using your bullets?2012-08-28
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    Henry Sir, can you make proof of (i) by using your bullets?2012-08-28
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    Sir, kindly look at the proof of first part. Becoz, I applied your bullets. But those bullets are not enough to complete the proofs. Please look. I am waiting for your solution.2012-08-28
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    @vidyaojal You will understand better why this leads to a proof if you experiment with various values of $A$ and use $a_0 = \lceil{A}\rceil - 1$ and $a_{i+1}=\left\lfloor{\frac{1}{A-A_i}}\right\rfloor +1 $, keeping track of $A-A_i$ and $\frac{1}{A-A_i}$2012-08-28
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    HENRY Sir, numerically you are very much correct. I checked with several examples. However, we need to prove mathematically instead of taking different values to A. So, find a proof and discuss. Thank you for your previous solution/hints. Kindly do the rest by giving a proof. please...2012-08-29
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    ! can you complete any one of the proof of A?please...2012-08-29
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    Henry Sir, can you complete any one of the proof of A?please...2012-08-29