Since y_n diverges towards infinity, there must exist an n_0 ∈ ℕ such that for all n ≥ n_0 : y_n ≥ c_n. (If it didn't exist, y_n wouldn't diverge to infinity since we could find an infinite subsequence of y_n which contains only values less than zero.)

Therefore:

lim(n -> ∞) x_n + y_n ≥ lim (n -> ∞) x_n + c_n = lim(n -> ∞) x_n + 0 = ∞

□

[–] Waraugh@lemmy.dbzer0.com 13 points 1 year ago (1 children)

So the answer is □?

[–] yetAnotherUser@feddit.de 9 points 1 year ago (1 children)

In case you aren't joking, '□' is used to indicate the end of a mathematical proof. It's equivalent to q.e.d

[–] Waraugh@lemmy.dbzer0.com 4 points 1 year ago (1 children)

I was not joking, which also probably explains why I have no idea what anything else in your post says.

[–] yetAnotherUser@feddit.de 4 points 1 year ago

No worries, I made the comment mostly for people with somewhat advanced knowledge in math. A year ago I wouldn't have understood any of it either.

[–] spacesweedkid27@lemmy.world 6 points 1 year ago

You beat me to it

[–] ssfckdt@mastodon.cloud 5 points 1 year ago* (last edited 1 year ago) (1 children)

i think this means that ∞ + ∞ > ∞

[–] yetAnotherUser@feddit.de 6 points 1 year ago (1 children)

Not quite. It's somewhat annoying to work with infinities, since they're not numbers. Technically speaking, ∞ + ∞ is asking the question: What is the result of adding any two infinite (real) sequences, both of which approaching infinity? My "proof" has shown: the result is greater than any one of the sequences by themselves -> therefore adding both sequences produces a new sequence, which also diverges to infinity. For example:

The series a_n = n diverges to infinity. a_1 = 1, a_2 = 2, a_1000 = 1000.

Therefore, lim(n -> a_n) = ∞

But a_n = 0.5n + 0.5n.

And lim(n -> ∞) 0.5n = ∞

So is lim(n -> ∞) a_n = 2 • lim(n -> ∞) 0.5n = 2 • ∞?

It doesn't make sense to treat this differently than ∞, does it?

[–] Voroxpete@sh.itjust.works 3 points 1 year ago (1 children)

Sounds like the infinite hotel paradox

[–] PipedLinkBot@feddit.rocks 2 points 1 year ago

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[+] 10_0@lemmy.ml 3 points 1 year ago* (last edited 1 month ago) (2 children)

[deleted]

[–] wieson@lemmy.world 4 points 1 year ago

2infinite 2furious

[–] yetAnotherUser@feddit.de 1 points 1 year ago

There are infinitely many natural numbers, right? 1, 2, 3, 4, 5, 6, ...., ∞ are all natural numbers.

We can now multiply all natural numbers with (-1) to get another sequence of infinite numbers: -1, -2, -3, -4, -5, -6, ..., -∞

So how many whole numbers are there? The whole numbers are those: -∞, ..., -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, ..., ∞
Looking at what we previously defined, would you say there are 2 * ∞ whole numbers? After all, it's the amount of natural numbers [∞] + the amount of natural numbers multiplied by (-1) [∞], which is the result of ∞ + ∞, isn't it?

But wait a second, are there really exactly infinite natural numbers? We could split the natural numbers into all even (2, 4, 6, 8, 10,...) and all uneven (1, 3, 5, 7, 9,...) numbers. But there are infinitely many even and infinitely many uneven numbers as well. So are there actually 2 * ∞ many natural numbers?`

BUT WAIT, what if we split the even numbers up even further? How about all numbers divisible by four (4, 8, 12, 16, 20, 24,...) and all numbers not divisible by four (6, 10, 14, 18, 22,...). Well, there are again infinitely many whole numbers divisible by four and infinitely many numbers not divisible by four. So are there 2 * ∞ many even numbers and therefore 3 * ∞ many natural numbers (2 * ∞ from even, 1 * ∞ from uneven)?

As you can see, we can divide one infinity into infinitely many infinities. Multiples of ∞ aren't really any meaningful - they are just ∞ again!

[–] Ataraxia@lemmy.world 1 points 1 year ago

16 minutes to live.