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If you add 1 + .5 + .25 + .05 to etc. will you ever get to 2?

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wolframalpha.com/input/?i=1+0.5+0.25+...
en.wikipedia.org/wiki/0.999...
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Yes

>.25
This makes sense.
>then .05
...Are you retarded?

I remember something like this. If you walk a mile one day, then half of that the second, and then half of that the third day, etc. Would you walk an infinite distance after infinite years?

1
.5
.25
.05

if you keep adding random numbers, eventually you'll get there

if you are trying to use the sequence X + 0.5*(2-X), starting with X=1, then you'll approach 2 but never reach it

whoops nigga

>1 + .5 + .25
Okay that's just adding 150% of the last term, it's exponential so it'll never reach 2
>.05
Nigger what are you doing

Nope, the added nubmber will get smaller with each division, so->1+.5+.25+.125+.0625=1.9375 will keep growing, but smaller. It will never be a 2

>150 percent
Why are you alive

wolframalpha.com/input/?i=1+0.5+0.25+...

Interesting, but if you add it infinitely, isn't it going to reach 2 eventually?

Yes, for sufficiently small values of 2.

Another user here. No never, 1.99999999999999999999999999999999999999999999999999975 is very close, it will keep getting closer, but its not 2

no you fucking retard, you will never reach 2. It will always be less. Stay in school bud

nice dubs. hope your life gets better.

Everyone who says "no" to this cannot into math.

It's the infinite sum of 1/2^n which equals 2.

nope. It's an exponential function.
it'll get close as shit, but it will never reach it.

(ill assume you meant half of 0.25 which is 0.125 not 0.05)
yes its called a series when you add numbers this way

this is a convergent series when -1 < r < 1

t1= first term (2 in thise case)
r= ratio (1/2 in this case)
you use this formula to to calculate the sum of every term from the first to the the infinite term

infinite sum = t1/(1-r)
=1/(1-0.5)
=1/0.5
=2

its infinite... every single term. its 2

What are you, 12?

Stop talking about things you don't understand...

lim (x -> inf) X + 0.5(2-X) = 2
x approaches but does not reach
first derivative test

...

It depends on how you look at it. The space between 1.999 repeating and 2 is infinitely small. So, 2 - 1.999... = 0, because something cannot be infinitely small. it's the same reason .99... and 1 are the same number.

So you got the infinite series
1/2^n1+1/2^n2+1/2^n3...
where the numbers after n, stand for the next number in the sequence of n. n is a whole number, and starts from 0->infinity+.
The sum of the infinite series will be 2, since it's convergent.

Yes, because no matter how slightly you move forward, you would still make infinite progress

OP, see this
Wolfram I suppose?

.999... = 1

Your point is moot.

Why reply to stuff you don't know? If you can't into math, just don't reply...

Yes, it's Wolfram because these people will actually argue about it forever if I just tell them how math works...

Nice, so because it's infinite, it actually does hit 2. Cool.

Wops, sorry, I thought you said it was not. It does not depend on anything though; it is 2. There is no other way to look at it..

not knowing the difference between approximately equal and actually equal

.99 is not 1. it's pretty damn close, but not equal

It is fucking equal you illiterate.

en.wikipedia.org/wiki/0.999...

>not knowing when you are out of your league

0.999 repeating is defined as 1.
In some other analysing method, it is defined to be an infinitesimal away from 1.
That is not standard analysis, so 0.999...=1

Theoretically

The limit is 2 but it will never be 2

>The limit
Okay retard

you wil be closer and closer to 2, but you'll never have it, it's like the perfect wife

This.

It is only "theoretically" because you can't into math and don't understand the concept of "infinite".

The sum of the CONVERGENT infinite series is equal to 2. Nothing to discuss.

>limit is 2 but it will never be 2

the difference will always be (0.5)(Xn), which will never be 0

Has no one here do basic limits?

It will tend to 2 but never reach.

thats not how it works. recurring digits are infinite, and infinity is never intuitive.

0.999... is not its own number, it is a different way of representing the number 1, and similarly the sum of 1/n to infinity is a representation of the number 2

We are doing it with a infinite geometric series, not with limits. So it is 2.

all of you saying this dont have a good grasp of what we're talking about. youre assuming you stop at some finite n, but you dont, its an infinite series and its not intuitive

dumb niggers

but what happens when i put it in my calculator?
i did it for a few minutes and it was 2?
idk can someone confirm this?
is my calculator broken?

>complaining about others not knowing limits
>get the wrong answer

Again, anything else than two is wrong, no matter "how you look at it.

Wolfram agrees with me and I have a university degree in this, so you are probably wrong...

Look for a proof online if you want, it's the infinite series 1/2^n from 0 to infinity..

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