Limit of a sequence does not goes to zero

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
2 replies · 1K views
Seydlitz
Messages
262
Reaction score
4

Homework Statement


Could you guys please verify this proof of mine? I want to show that a limit with particular property does not go to 0. It is part of the proof that when a sequence have an ever increasing term then the limit of the sequence is not 0.

The Attempt at a Solution



The ##\lim_{n \to \infty} a_n \neq 0## if ##|a_n|<|a_{n+1}|##

Suppose ##\lim_{n \to \infty} a_n = 0##, then there exist ##n>0##, such that ##|a_N| < \epsilon## when ##N>n##. Taking an arbitrary ##a_n## as ##\epsilon##, we can get ##|a_N| < |a_n|.## Because it also true that ##N=n+1>n##, we get ##|a_n|>|a_{n+1}|##. A contradiction.

Hence it is not possible for the limit to be 0.

Thank You
 
Physics news on Phys.org
Seydlitz said:

Homework Statement


Could you guys please verify this proof of mine? I want to show that a limit with particular property does not go to 0. It is part of the proof that when a sequence have an ever increasing term then the limit of the sequence is not 0.

The Attempt at a Solution



The ##\lim_{n \to \infty} a_n \neq 0## if ##|a_n|<|a_{n+1}|##

Suppose ##\lim_{n \to \infty} a_n = 0##, then there exist ##n>0##, such that ##|a_N| < \epsilon## when ##N>n##. Taking an arbitrary ##a_n## as ##\epsilon##, we can get ##|a_N| < |a_n|.## Because it also true that ##N=n+1>n##, we get ##|a_n|>|a_{n+1}|##. A contradiction.

Hence it is not possible for the limit to be 0.

Thank You

That looks just fine to me.
 
  • Like
Likes   Reactions: 1 person
Ok thanks for your verification Dick! I'm quite happy to discover this by myself.