Bijection of an infinite series

**osudude** · November 30th 2009, 08:51 PM

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
following:

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.

**Enrique2** · December 1st 2009, 12:17 AM

What you are asking is exactly how to prove a particular case of the Riemann rearrengement theorem, that is if a series $\sum_n a_n$ is convergent but the series of its abolute values diverge then, for each $A\in\mathbb{R}$ there exists a permutation $\pi:\mathbb{N}\rightarrow\mathbb{N}$ such that $\sum_na_{\pi(n)}=A$ .

In this especific case, $a_n=\frac{(-1)^n}{n}$ , you have that both $\sum_{n}a_{2n}$ and $\sum_{n}a_{2n-1}$ are divergent (in the general case you would have to take the positive and the negative terms). Now, if $A$ is positive choose $n_0$ such that $P_1:=\sum_{n=1}^{n_0}a_{2n}>A$ . Now define $\pi(n)=2n$ for $n\leq n_0$ . Again since $\sum_{n}a_{2n-1}$ is divergent to $-\infty$ you can take $n_1$ such that $N_1:=P_1-\sum_{n=1}^{n_1}a_{2n-1}<A$ . Take $\pi(n_0+n)=2n-1$ for $n=1,\ldots,n_1$ . Now you use
that $\sum_{n=n_0+1}^{\infty} a_{2n}$ and
$\sum_{n=n_1+1}^{\infty} a_{2n+1}$ are divergent to "move after A and before A" recursively. The key to show that the proces leads is that $a_n$ 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.

**HallsofIvy** · December 1st 2009, 03:34 AM

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

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