# given a real number, build an infinite series that will converge to it

I wonder if there's a method for building a series of real numbers that will converge to a given real number thanks

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Is there any other way to be given a real number? –  user1708 Mar 17 '12 at 18:19
For a given real number $c$ take the series $\sum a_k$ with $a_0 = c$ and $a_k = 0$ for $k > 0$. –  Rudy the Reindeer Mar 17 '12 at 18:19
That's interesting, at this moment one downvote but seven answers... –  Dirk Mar 17 '12 at 21:00
Matt, your type of humor is partially why I visit this site. Nice answer. –  000 Mar 17 '12 at 21:03
@Dirk: Does the question show any research effort? And, as Holowitz mentions, the question is (kind of) contradictory. –  TMM Mar 17 '12 at 21:30

Take any series whose sum diverges and terms go to zero (e.g., $1/n$). Keep adding terms until the sum is greater than or equal to the desired number (so you don't get equality), throw away the last term added, and continue until the cheese is melted or an infinite number of terms has been added (take with a grain of salt).

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simple and useful - thanks! –  hershalle Apr 2 '12 at 11:40

If you are looking for a sequence of rationals converging to the real number $r$, here is one way $\displaystyle q_n = \frac {\left\lfloor {10^n r} \right \rfloor}{10^n}$

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There are many such techniques. For example, take successive approximations to its decimal representation: Given $\pi$, define $a_1 = 3$, $a_2 = 3.1$, $a_3 = 3.14$ etc. Of course, this requires that you know the decimal representation.

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thanks! good idea, but good for only specific cases –  hershalle Apr 2 '12 at 11:38

Of course we can build a series of real numbers that converges to any real. (However, the way I show is a trivial construction, perhaps you are looking for something more interesting). We can take any series which we already know converges, and manipulate it slightly. For example:

Let $r\in\mathbb{R}$.

We know that the sum $$\sum^{\infty}_{1}\frac{1}{n^2} = \pi^2/6$$ (if you don't, it can be shown simply with parseval's equality.)

So then $$\sum^{\infty}_{1}\frac{6r}{\pi^2n^2} = r$$

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Take any convergent sequence $\sum a_n=c$. If your number is $a$, then you can take $\sum b_n$ where $b_0=a_0-c+a, b_i=a_i (i>0)$

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Alternatively, adding/subtracting $2$ if necessary, you could express it as a geometric series. For example, $\pi$.

$\pi = \dfrac{1}{1-x}$ for some $x$ with $|x| < 1$. In particular, $x = \dfrac{\pi - 1}{\pi}$. But then the series $1 + x + x^2 + ...$ converges to $\pi$.

Yet another - if we can compare the real number with dyadics easily, then you can iteratively find the largest $c_i$ such that $c_0 + c_1/2 + c_2/2^2 + \ldots$ converges to the real, simply by subtracting the real from the $(i-1)^\text{st}$ expansion and seeing if it's greater or less tha $1/2^i$

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