Real number in terms of infinite series

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$$\ge1 and 2 \ge a_i for i\ge2 with a_i = 2. Again, here k \ge1 and k is some fised real number. Can you generalize with an example of this modified series? - What does k add? From the looks of it I can always choose k=1 and just pick my constants differently to "simulate" any power of some other series that I like. – Robert Mastragostino Aug 30 '12 at 5:55 @Robert Mastragostino! not exactly k =1. here the statement is valid for all k > or = 1. Of course, k is fixed real number. – vidyaojal Aug 30 '12 at 6:45 Maybe you could link to your previous post? – celtschk Aug 30 '12 at 6:50 @celtschk! you can see the link: math.stackexchange.com/questions/187774/… – vidyaojal Aug 30 '12 at 7:10 @vidyaojal: Thank you. I've edited it into the question, so that others don't need the comments to find out. – celtschk Aug 30 '12 at 7:12 2 Answers Given that from your condition it is allowed that a_{i+1}=a_i, if k is integer, you can reduce the problem for k>1 to the case k=1 by repeating 1/a_i^k for a_i^{k-1} times before using a new value. By doing that repetition, you get a sum of a_i^{k-1}/a_i^k = 1/a_i. Example: Say your k=1 expansion is$$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$1<k<2$. - Although in the comments above, the poster has stated that$k$is a fixed real number, not necessarily an integer. – Old John Aug 30 '12 at 7:34 @Old John: I didn't notice that. For me, the name choice of$k$together with it being an exponent suggested that it's an integer (well, strictly speaking, he also didn't specify that the$a_i$are integers, therefore one could just choose$a_0$as appropriate real number, no matter what the rest of the series converges to ... – celtschk Aug 30 '12 at 7:40 I agree that the question needs to be defined much more clearly! – Old John Aug 30 '12 at 7:43 @vidyaojal: I've not just given an example, I've added an example for the general rule for integer$k>1$I've given (that is, I've generalized it for all integer$k$with$k>1$), and I also said that for non-integer$k$the same strategy can be used to reduce the prtoblem to the (probably simpler) case$1<k<2$(this case I didn't solve, however). That is, I've solved the problem completely for integer$k$and gave a possible step in solving it for non-integer$k$. That's definitely more than just giving an example. I admit it is not a complete solution for your problem, though. – celtschk Aug 30 '12 at 8:03 @vidyaojal: BTW, I just now notice that with the conditions as given, no$A<2$can be written that way (because your condition$a_i\ge 2$doesn't exclude$a_0\$) – celtschk Aug 30 '12 at 8:08

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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! you are right and I studied the same after your post. Can you look my series and generalize? – vidyaojal Aug 30 '12 at 9:10
! can you apply the same for my post? – vidyaojal Aug 30 '12 at 9:20
Koarik! r u there? Please see my new edited post and respond strictly on it. – vidyaojal Aug 30 '12 at 9:41