Proof: c * divergent sequence diverges

tarheelborn
Messages
121
Reaction score
0

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.

Homework Equations





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.
 
Physics news on Phys.org
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 \rightarrow L we will show that an \rightarrow L/|c|, which is a contradiction.

Suppose\ \epsilon &gt; 0. Then there is N > 0 such that:

| can - L| < |c|\epsilon

for all n > N. Therefore

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

which gives, upon dividing that last inequality by |c|, | an - L/c| < \epsilon
for all n > N.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top