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Thread: Bijection of an infinite series

  1. #1
    Nov 2009

    Bijection of an infinite series

    So here is the question, that I am just really stuck on :

    Problem. Let x(n) = (-1)^n / n, and A be a real number. Prove the

    there exists such a 1-to-1 mapping (bijection) p : N --> N that
    an infinite series with a generic term x(p(n)) converges and the sum of
    this series is equal to A .

    ok, So I understand that as n approaches infiniti, this series is really getting to zero from both sides. understood. so we need to prove it's an injection and a surjection. for an injection, do you just have to prove the basic fact that if f(x_1) = f(x_2), then x_1=x_2 ?? Now with the surjection, I am just confused in general, and don't even know how to approach it.

    Now with this term A, we want to prove that the series converges (so would i just show the limit, as stated above? or how would i do this with a real analysis approach??) and also, the sum of this series is what? how can you conclude that its sum is A??

    Thank you for the help. Just wanted to make clear what I understand and don't understand, etc.
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  2. #2
    Jun 2009
    What you are asking is exactly how to prove a particular case of the Riemann rearrengement theorem, that is if a series $\displaystyle \sum_n a_n $ is convergent but the series of its abolute values diverge then, for each $\displaystyle A\in\mathbb{R}$ there exists a permutation $\displaystyle \pi:\mathbb{N}\rightarrow\mathbb{N}$ such that $\displaystyle \sum_na_{\pi(n)}=A$.

    In this especific case, $\displaystyle a_n=\frac{(-1)^n}{n}$, you have that both $\displaystyle \sum_{n}a_{2n}$ and $\displaystyle \sum_{n}a_{2n-1}$ are divergent (in the general case you would have to take the positive and the negative terms). Now, if $\displaystyle A$ is positive choose $\displaystyle n_0$ such that $\displaystyle P_1:=\sum_{n=1}^{n_0}a_{2n}>A$. Now define $\displaystyle \pi(n)=2n$ for $\displaystyle n\leq n_0$. Again since $\displaystyle \sum_{n}a_{2n-1}$ is divergent to $\displaystyle -\infty$ you can take $\displaystyle n_1$ such that $\displaystyle N_1:=P_1-\sum_{n=1}^{n_1}a_{2n-1}<A$. Take $\displaystyle \pi(n_0+n)=2n-1$ for $\displaystyle n=1,\ldots,n_1$. Now you use
    that $\displaystyle \sum_{n=n_0+1}^{\infty} a_{2n}$ and
    $\displaystyle \sum_{n=n_1+1}^{\infty} a_{2n+1}$ are divergent to "move after A and before A" recursively. The key to show that the proces leads is that $\displaystyle a_n $ goes to 0, then passing to before A to after A or the converse will be "with a such small step as desired" for a sufficient number of repetitions.

    This is just a sketch of Riemann adapted to your case, the idea is simple, but writting with precision is a little bit harder.
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  3. #3
    MHF Contributor

    Apr 2005
    Is that really what the problem says, just find a bijection from $\displaystyle \{\frac{(-1)^n}{n}\}$ to a series that converges to A?

    Here is what I would do: First find a series that converges to a. Since the geometric series $\displaystyle \sum_{n=0}^\infty r^n$ converges to $\displaystyle \frac{1}{1-r}$, set $\displaystyle \frac{1}{1- r}= A$ and solve for r:
    $\displaystyle r= \frac{A-1}{A}$ so the series $\displaystyle \sum_{n=0}^\infty \left(\frac{A-1}{A}\right)^n$ converges to A. Now just do the obvious bijection: map $\displaystyle \frac{(-1)^n}{n}$ to $\displaystyle \left(\frac{A-1}{A}\right)^n$. Surely, there is more to it than that!
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