DAILY MATH CHALLENGE THREAD #4

This is easily done using the closed form for the fibonacci sequence.

Let's try for a solution that doesn't use that.

1/3?

Just to clarify:
F_1 = 1
F_2 = 1
F_3 = 2
?

>>8672552
It is? By closed form you mean (Phi^n - phi^n) / sqrt(5)?

>>8672556
Nope

>>8672561
Yeah, 1, 1, 2, 3, 5, etc

>>8672570
>Nope
whoops, 4/15?

>>8672570
Yes, given that closed form you just break the given sum into two geometric series. I mean, it's a bit messy but conceptually easy.

>>8672549

[eqn]\sum_{n=1}^\infty \frac{F_{n}}{5^n} = \sum_{n=1}^\infty \frac{\phi^n - \psi ^n}{\sqrt{5} *5^n} = \frac{1}{\sqrt{5}} \left( \sum_{n = 1}^\infty \left(\frac{\phi}{5} \right)^n - \sum_{n=1}^{\infty} \left( \frac{\psi}{5} \right)^{n} \right)[/eqn]

I hope it compiles well, it did in the preview. At the last step, you just calculate the geometric series and whatever it gives is the result.

Sorry if someone has already solved it.

If you want a nice topology problem, proof that in the real kolmogorov line, a set A is compact iff:

1.- You can put the set inside a set of the form [a, b)
2.- A is closed
3.- There's no strictly increasing sequence inside of A

Another version asks to prove the same but instead of conditions 1 and 2 you have that any strictly decreasing sequence inside of A eventually converges in A.

>>8672581
reee im bad at algebra, 5/19

>>8672570
>Yeah, 1, 1, 2, 3, 5, etc

In that case, we have
[eqn]S = \sum_{n \ge 1} \frac{F_n}{5^n} = \frac15 +\frac{1}{25} + \sum_{n \ge 3} \frac{F_{n-2} + F_{n-1}}{5^n} [/eqn]
and most likely (Im too lazy) you just write the last expression in terms of S again and then you can solve for S. (this is valid as long as we assume the series actually converges)

>>8672584
>>8672588
Oh I see, there's a much nicer solution though.

>>8672591
Correct!

>>8672604
Yeah that's the idea

[eqn]S = \sum_{n = 1}^{\infty} \frac{F_n}{5^n} = \frac{1}{5} + \frac{1}{25} + \frac{2}{125} + \frac{3}{625} + ...[/eqn]
[eqn]\frac{S}{5} = \frac{1}{25} + \frac{1}{125} + \frac{2}{625} + \frac{3}{3125} ...[/eqn]
[eqn]\frac{4S}{5} = \frac{1}{5} + \frac{1}{125} + \frac{1}{625} + \frac{2}{3125} + ...[/eqn]
[eqn]\frac{4S}{5} = \frac{1}{5} + \frac{S}{25}[/eqn]
[eqn]\frac{19S}{25} = \frac{1}{5}[/eqn]
[eqn]S = \frac{5}{19}[/eqn]

File: fib sum work.jpg (36KB, 476x653px) Image search: [Google]

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Here's my work

I guess its fairly obvious the series converges. For example, by induction we have that F_n is at most say 2^n. So this method of "solving for S" is valid since S is a finite real number

b) Evaluate [eqn]\sum_{n=1}^\infty \frac{F_n}{N^n}[/eqn]

>>8672625
Is that supposed to be a lowercase "n" on the bottom? If so I doubt this one has any kind of nice answer

define the predecessor function in the untyped lambda calculus

>>8672637

Is there even a closed form for [eqn]\sum_{n=1} ^{\infty} \frac{1}{n^{n}}[/eqn]

>>8672672
sort of
https://en.wikipedia.org/wiki/Sophomore's_dream

Here's a kinda stupid one, but hopefully one of you finds it amusing. Let X be a non-empty set. Show that X admits a poset structure with a greatest element without using AC

for you, my children

File: Screen Shot 2017-02-13 at 5.24.35 PM.png (48KB, 673x239px) Image search: [Google]

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>>8672718
crap forgot photo

File: 1486710405202.jpg (22KB, 368x285px) Image search: [Google]

22KB, 368x285px

>>8672721

> A combinatorial problem

I wonder why they're so popular in these threads.