Proving a sequence converges using the formal definition  Series  AP Calculus BC  Khan Academy  Summary and Q&A
TL;DR
This video provides a proof that a specific sequence converges to 0.
Questions & Answers
Q: What is the specific sequence being analyzed in this video?
The sequence being examined in the video is (1)^n+1/n.
Q: What does it mean for a sequence to converge?
Convergence means that as the index of the sequence approaches infinity, the terms of the sequence get arbitrarily close to a specific value or limit.
Q: How is epsilon used in proving the convergence of the sequence?
Epsilon is used to define a range around the limit value. The proof shows that, for any epsilon greater than 0, there exists an M such that if n is greater than M, the sequence will be within epsilon of the limit.
Q: How is the value of M determined in the proof?
The value of M is determined by taking the reciprocal of epsilon. Thus, M is set as 1/epsilon to ensure that for n greater than M, the sequence is within epsilon of the limit.
Q: Why does the sequence (1)^n+1/n converge to 0?
The proof shows that for any given epsilon, there exists an M such that if n is greater than M, the sequence is within epsilon of 0. Therefore, as the index of the sequence increases, the terms get arbitrarily close to 0.
Summary & Key Takeaways

The video presents the claim that a sequence defined as (1)^n+1/n converges to 0 but lacks proof.

To prove the convergence, the video introduces the concept of epsilon and demonstrates how to find a value of M that ensures the sequence is within epsilon of 0.

By taking the reciprocal of both sides of the inequality, the video shows that the sequence converges when n is greater than 1/epsilon.

The proof is valid for any epsilon greater than 0, demonstrating that the limit of the sequence is indeed 0.