Math Is Fun Forum

Multiply each side of the equation BP/PA = QC/AQ by AQ/QC.

I guess you could say the same thing about many holidays... the government invented them to steal our money of course

I have four of them

We have two sequences of real numbers {a[sub]n[/sub]} and {b[sub]n[/sub]} such that {a[sub]n[/sub]} is a decreasing positive sequence (a[sub]n[/sub] ≥ a[sub]n+1[/sub] > 0) which converges to 0 and the sum of countably many terms of {b[sub]n[/sub]} is bounded (|∑b[sub]n[/sub]| ≤ M for some constant M). Then

converges.

I will use the problem you posted in your other thread to give a simple example:

Let {a[sub]n[/sub]} = 1/n and {b[sub]n[/sub]} = sin n. Clearly, 1/n ≥ 1/(n + 1) > 0 and lim(1/n) = 0, so that {a[sub]n[/sub]} satisfies the conditions of Dirichlet's test. Since |sin x| ≤ 1 for any real x, we have that

for every positive integer N. Then {b[sub]n[/sub]} satisfies the conditions for Dirichlet's test. Thus

converges.

That is a good point to bring up George; it seems that we have merely assumed the limit exists and have not addressed the fact that the sequence may "oscillate" too much to converge. So let us determine if {a[sub]n[/sub]} is convergent. Clearly, {a[sub]n[/sub]} is bounded, since for any x ∈ R, -1 ≤ cos x ≤ 1, and thus |a[sub]n[/sub]| ≤ sup{1, |a[sub]0[/sub]|}. Then by the Bolzano-Weierstrass Theorem, {a[sub]n[/sub]} has a convergent subsequence. Now I leave the rest of this to someone else, and perhaps I will post a solution soon if no one seems to want to step up to the plate: {a[sub]n[/sub]} is convergent if every convergent subsequence of {a[sub]n[/sub]} has the same limit. Can you prove that this sequence has that property? I am very tired so I may be giving a difficult approach to the problem, but I am confident that it does in fact converge.

If we assume that the sequence does converge, then Ricky's solution is valid, since for any convergent sequence {a[sub]n[/sub]}, any tail of the sequence is also convergent to the same limit; in particular, lim(a[sub]n[/sub]) = lim(a[sub]n+1[/sub]).

Yes, Dross provided the solution in the thread you posted in earlier...

http://www.mathsisfun.com/forum/viewtop … 144#p57144

Theoretical physicist and mathematician.

Brute force might take you a while, write some ideas on what you can say about numbers with this property down on paper, and maybe some ideas will come to you.

The member who posted about the "all problem solving algorithm" was se7en. This is the guy you're talking about, correct?

http://www.mathsisfun.com/forum/viewtopic.php?id=4028

He deleted most of what he said, but you'll probably remember the thread if we are thinking of the same guy.

I tried for a long time to get "butts" to show up in the list

Here is a "find the function" problem that a friend of mine came up with:

Let f(x) = 1 - 2x^2. Find a function g(x) such that g(g(x)) = f(x) for all x, or prove no such function exists.

It is way kl.