Proof: c * divergent sequence diverges

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SUMMARY

The discussion centers on proving that if {a_n} is a divergent sequence of real numbers and c is a non-zero real number, then the sequence {c*a_n} also diverges. The proof employs a contradiction approach, demonstrating that if {c*a_n} were convergent, it would imply that {a_n} converges, which contradicts the initial condition of divergence. The proof is structured around the definition of limits and the properties of absolute values, ultimately confirming the divergence of the scaled sequence.

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Homework Statement



Suppose that {a_n} is a divergent sequence of real numbers and c \in R, c <> 0. Prove that {c*a_n} diverges.

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The Attempt at a Solution

I have attempted to solve the problem as a proof by contradiction, but am afraid I am leaving something out. Please confirm my proof is complete or prompt me to add. Thanks!

Proof is by contradiction. Suppose {c*a_n} is convergent. This means that |c*a_n - L| < e, for e > 0. Then there is N \in N such that |a_n - L/c| < e/|c|, n >= N. But this is the definition of limit of a sequence, so a_n converges. But this contradicts our problem statement so, in fact, a_n diverges. End of proof.
 
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You have the right idea, but it isn't written well. You need to use your observations to write your argument in reverse. Something like this:

Given can [itex]\rightarrow[/itex] L we will show that an [itex]\rightarrow[/itex] L/|c|, which is a contradiction.

Suppose[itex]\ \epsilon > 0[/itex]. Then there is N > 0 such that:

| can - L| < |c|[itex]\epsilon[/itex]

for all n > N. Therefore

|can - L| = |c(an - L/c| = |c||(an - L/c| < |c|[itex]\epsilon[/itex]

which gives, upon dividing that last inequality by |c|, | an - L/c| < [itex]\epsilon[/itex]
for all n > N.
 

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