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 = ∞

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[–] 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 (2 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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[–] ssfckdt@mastodon.cloud 0 points 1 year ago (1 children)

Wait, isn't there some thought experiment where you can insert infinity into infinity simply by moving infinity over by one infinite times?

I'm too lazy to look it up rn

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

Yup, someone else commented it in this thread.

https://sh.itjust.works/comment/3777415

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