# {Sn} is convergent -> {|Sn|} is convergent

• Jamin2112
In summary, the conversation discusses proving that if {sn} is convergent, then {|sn|} is convergent by using the contrapositive method. The goal is to show that ||Sn|-|S|| <= |Sn-S|, which can be done by using the triangle inequality. Another problem is also discussed, which involves proving that if a subsequence of a Cauchy sequence converges, then the full sequence also converges. The hint for this problem is to use contradiction and the fact that the subsequence is also a Cauchy sequence.

#### Jamin2112

{Sn} is convergent ---> {|Sn|} is convergent

## Homework Statement

I need to prove that if {sn} is convergent, then {|sn|} is convergent.

## Homework Equations

sn is convergent if for some s and all ε > 0 there exists a positive integer N such that |sn - s| < ε whenever nN.

## The Attempt at a Solution

Proof. By contrapositive. Suppose {|sn|} is not convergent. Then for all s there exists an ε > 0 such that ||sn| - s|| ≥ ε for all n.

... I need to somehow show that this implies that {sn} does not converge.

Maybe some fancy triangle inequality thing like

ε ≤ ||sn| - s| ≤ ||sn| - sn| + |sn - s|

Wat do, PF?

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Ah, wait. I think I figured out something.

ε ≤ ||sn| - s| ≤ ||sn| - sn| + |sn - s| ≤ 2|sn| + |sn - s|

Am I any closer?

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You are supposed to show that |Sn| converges to |S| not to S.
The trick is to show that ||Sn|-|S|| <= |Sn-S|

Hint to showing it:
|S| = |Sn + (S-Sn)|
And use the triangle inequality.

MaxManus said:
You are supposed to show that |Sn| converges to |S| not to S.
The trick is to show that ||Sn|-|S|| <= |Sn-S|

Hint to showing it:
|S| = |Sn + (S-Sn)|
And use the triangle inequality.

I knew the triangle inequality would show his face somewhere. http://images.icanhascheezburger.com/completestore/2009/3/5/128807888821558038.jpg

Hey, I need a hint for another problem. (You guys give stellar hints. Pat yourselves on the back.)

Suppose {pn} is a Cauchy sequence in a metric space X, and some subsequence {pni} converges to a point p in X. Prove that the full sequence {pn} converges to p.

As always, I'd love to use contradiction.

Suppose {pn} does not converge to p. Then for some ε > 0, d(p,pn) ≥ ε for all n. Then d(pm,pn)+d(pm,p) ≥ ε for all m, n. But {pm} is a Cauchy sequence, so with the proper choice of m, n we'll have d(pm,pn) < ε. ... blah blah I hope this is going somewhere. A little hint, maybe?