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This is just continuation of my previous post.

$ A = a_0 + \left(\frac{1}{a_1}\right)^k + \left(\frac{1}{a_2}\right)^k + \left(\frac{1}{a_3}\right)^k +\ldots$ Where $i\ge1$ and the recurrence relation $a_{i+1}\ge a_i\ge2$.

I just raised for each term by $k$. here $k\ge1$ and $A$ is any given real number. Then the above series will exist. Can we generalize this series of some real number with those initial conditions? If yes, kindly discuss...

edit The first term is $a_0$ but not $a_1$. Of course I edited now.

Edited and extended $ A = a_0 + \left(\frac{1}{a_1}\right)^k + \left(\frac{1}{a_1}\right)^k \left(\frac{1}{a_2}\right)^k + \left(\frac{1}{a_1}\right)^k \left(\frac{1}{a_2}\right)^k\left(\frac{1}{a_3}\right)^k +\ldots$

Where $a_1$ = 2, $a_i$$\ge1$ and 2 $\ge$ $a_i$ for $i\ge2$ with $a_i$ = 2. Again, here k $\ge1$ and k is some fised real number. Can you generalize with an example of this modified series?

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    @vidyaojal: Thank you. I've edited it into the question, so that others don't need the comments to find out.2012-08-30

2 Answers 2

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Given that from your condition it is allowed that $a_{i+1}=a_i$, if $k$ is integer, you can reduce the problem for $k>1$ to the case $k=1$ by repeating $1/a_i^k$ for $a_i^{k-1}$ times before using a new value. By doing that repetition, you get a sum of $a_i^{k-1}/a_i^k = 1/a_i$.

Example:

Say your $k=1$ expansion is $A = 1 + 1/2 + 1/8 + \dots$ and $k=2$. Then your new expansion reads $\begin{align} A &= 1 + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2\\ &\phantom{= 1} + \left(\frac{1}{8}\right)^2 + \left(\frac{1}{8}\right)^2 + \left(\frac{1}{8}\right)^2 + \left(\frac{1}{8}\right)^2\\ &\phantom{= 1} + \left(\frac{1}{8}\right)^2 + \left(\frac{1}{8}\right)^2 + \left(\frac{1}{8}\right)^2 + \left(\frac{1}{8}\right)^2 +\dots \end{align}$

Since every real number can be written as the product of an integer and a real number in the range $[1,2)$, the same strategy can be used to reduce the problem for arbitrary non-integer $k>1$ to the problem $1.

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    @celtschk! Are you there Sir? Please see my new edited post and respond strictly on it.2012-08-30
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If I am interpreting your question and intent correctly -- which is probably not the case -- then yes. In fact, all real numbers can be defined in terms of (Cauchy-) converging sequences of rational numbers.

I am sure that you already know that a series is a just sequence of partial sums so that, in a given base, the "power-series" representation of a number is just a converging sequence of rational numbers. For example, the decimal expansion of pi is 3.1459... . We could view this as the limit of the rational sequence {3, 3.1, 3.14, 3.145, ...}.

A classic way to define a real number is by a sequence of rational numbers whose successive differences converge to zero ("Cauchy convergence").

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    Koarik! r u there? Please see my new edited post and respond strictly on it.2012-08-30