The discussion focuses on proving that the sequence √a(n) converges to 0, given that a(n) converges to 0. The proof begins by establishing that for any ε > 0, there exists an n0 in the natural numbers such that for all n ≥ n0, √a(n) < ε. Participants emphasize the need to demonstrate that √a(n) can be made less than specific values, such as 1/100 and 1/1000000, as n approaches infinity.
PREREQUISITESMathematics students, educators, and anyone involved in analysis or proof-based mathematics who seeks to understand convergence and its implications in sequences.
The rest of your thought isCoachBryan said:I've been messing with this proof for while and I'm stuck on this. I've started with a(n) converges to 0, let epsilon > 0, then there exists an n0 in N such that for all n >= n0.
I'm stuck here thus far. Any help? Thanks for your time.
Mark44 said:The rest of your thought is
For all n >= n0, ##\sqrt{a_n} < \epsilon##.
What are the given conditions? Is it an converges to 0? You have an = 0 in the title.