/sci/ - Science & Math

[Anonymous No.7906524]

Can someone prove this?

[Anonymous No.7906527]

>>7906524
Homework threads >>>/hm/

[Anonymous No.7906546]

>>7906524
Easy. 2+2+2+....=-1/12 <2, so by reverse squeeze corollary, qed.

[Anonymous No.7906547]

Plug it into your calculator.

[Anonymous No.7906548]

>>7906546
Forgot to mention lhs<-1/12

[Anonymous No.7906553]

>>7906524
Why don't you calculate it you faggot?
Literally up to [math]10^{-100}[/math] error, or even less.

[Anonymous No.7906562]

>>7906553
I need prove, i don't need answer

[Anonymous No.7906563]

>>7906524
sum=0
for i=2016:-1:1
sum=(sum+i)^(1/i);
end
sum
=1.91163921625

[Anonymous No.7906571]

>>7906562
>I need prove a stupid algebraic equation
you brought this upon yourself you stupid mathfag.

[Anonymous No.7906585]

>>7906524

[math] 1 < \sqrt[n]{n} < 2 [/math]
[math] 1 < \sqrt[n]{n+2} ≤ 2 [/math]

[Anonymous No.7906594]

>>7906524
given f(x)=(x)^(1/x) notice f(x>2)<f(2)=sqrt(2)
then sqrt(2+sqrt(2+sqrt(2+...))) composed 2015 times is greater than LHS.
The infinite expression sqrt(2+sqrt(2+...)) equals 2
The finite expression 2015 times is less than the infinite one, therefore less than 2

[Anonymous No.7906599]

>>7906594
samefag, made a mistake (just checked with WolframAlpha)

f(x)=(x)^(1/x) has max at x=e, not x=2

so same approach, but do cuberoot{3+cuberoot{3+...}}
That has infinite sum x=1.67ish

Then apply final sqrt(2+1.67) < 2

[Anonymous No.7906607]

>>7906524
square both sides
subtract 2
cube both sides
subtract 3
keep doing this on the other side you'll have

2^2-2=2
2^3-3=5
5^4-4=621

keep doing this and u get something fuck huge on the other side, 2016 on the other

[Anonymous No.7906610]

>literally 'the current year' of gimmick number theory problems

[Anonymous No.7906645]

>>7906607
Thank you, you are great

[Anonymous No.7906657]

>>7906610
the AIME was yesterday. Too bad you never qualified

[Anonymous No.7906683]

>>7906585
the nth root of n+2 is not always less than 2 faggot, its only true for n >= 3

[Anonymous No.7906696]

>>7906657
>AIME
should I know what that is?

[Anonymous No.7906717]

>>7906696
American Invitational Mathematics Examination
> series of exams with cutoffs to form the team that competes at International Mathematics Olympiad
> Last year's results
> 1 usa
> 2 china
> 3 south korea
> 4 north korea
wat

[Anonymous No.7906738]

>>7906683
>retarded CS major that doesn't know what ≤ means detected

[Anonymous No.7906971]

>>7906562
>stupid mods delete my answer because they're butthurt mathfags
>they don't delete homework threads

/sci/ confirmed for absolute shit once again.

[Anonymous No.7907030]

>>7906524
Since [math] \sqrt [m] { n } \leq \sqrt { n } ~ \forall ~ n \geq 2 [/math] then you can solve the simpler problem of nested square roots, ie [math] \sqrt { 2 + \sqrt [ 3 ] { 3 + \cdots } } < \sqrt { 2 + \sqrt { 3 + \cdots } } \leq 2 [/math]

>>7906971
Shame they didn't ban your faggot ass.

[Anonymous No.7907796]

>>7907030
That's not how you use universal quantifiers.
>first year detected

[Anonymous No.7907803]

>>7907030
You're absolutely wrong and your notation sucks, get out.

[Anonymous No.7907874]

>>7907030
That equals 2.09

>>7907796
There's nothing wrong with writing the for all after the expression.