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Sequences

I've been asked to find a sequence such that Sn is divergent but |Sn| is convergent. I can't quite figure out a sequence that satisfies this; I thought of Sn = (-1)^n so that |Sn| = 1 but (-1)^n can't be considered divergent, can it?
Original post by r3l3ntl3ss
I've been asked to find a sequence such that Sn is divergent but |Sn| is convergent. I can't quite figure out a sequence that satisfies this; I thought of Sn = (-1)^n so that |Sn| = 1 but (-1)^n can't be considered divergent, can it?


Since a divergent sequence is one that isn't convergent, then yes, that's fine. It doesn't have to go zooming off to infinities, just not converge.
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Original post by ghostwalker
Since a divergent sequence is one that isn't convergent, then yes, that's fine. It doesn't have to go zooming off to infinities, just not converge.


ah I see. Thanks!
Original post by r3l3ntl3ss
ah I see. Thanks!


Yeah that is a good answer. Intuitively you can see that the sequence sn=(-1)^n would diverge by supposing that it does converge but then it has a distinct limit L say. But what could L possibly be since the sequence takes a different value for each successive n then it is clear (and fairly straightforward) to prove it has no limit. I.e it diverges. :smile:
Original post by poorform
Yeah that is a good answer. Intuitively you can see that the sequence sn=(-1)^n would diverge by supposing that it does converge but then it has a distinct limit L say. But what could L possibly be since the sequence takes a different value for each successive n then it is clear (and fairly straightforward) to prove it has no limit. I.e it diverges. :smile:
Or use:

Converges => Cauchy. But ∣an+1−an∣=2|a_{n+1}-a_n| = 2 so sequence can't be Cauchy.
Original post by DFranklin
Or use:

Converges => Cauchy. But ∣an+1−an∣=2|a_{n+1}-a_n| = 2 so sequence can't be Cauchy.


That is a very nice way to prove it imo much more quicker than my way. Although it does rely on knowing Cauchy sequences which OP may not have covered (yet).

Thanks for sharing.

:smile:
Alternatively, taking ε=1\varepsilon = 1 in the definition of convergence should get you an easy contradiction to it being convergent.

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