Proving lim(1/n) = 0

Real Analysis · Axiom Academy

A complete –N proof, built straight from the definition of a limit. Prove, directly from the definition of the limit of a sequence, that . That is: for every we must produce an index N so that holds for all . Here is the proof at a glance for . The terms approach 0 from the right. The green strip is the tolerance band . Since , we take N=6 : every term with (green) has landed inside the band, while the earlier terms (red) have not. Read it off: shrink and the band narrows, pushing the threshold N further right — but there is always such an N , because keeps decreasing toward 0 . That is exactly what means. Nice work — you gave a complete – N proof that , the template every limit proof in this course follows. Start from the definition: the limit is 0 exactly when, for every , all but finitely many terms satisfy . Work the inequality: for n > 0 , , and . Name an explicit N : any integer works, e.g. . Then . Verification is the proof: producing N and checking the tail is what makes the limit rigorous — not just watching the terms shrink. The pattern — unwind the absolute value, solve for n , choose N past the cutoff, verify the tail — is the engine behind every sequence-limit proof you will write. The hard part is rarely the algebra; it is committing to an explicit N .

This is the written version of the interactive lesson above. See the full Real Analysis course.