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Proof

(related to Proposition: Limit of 1/n)

• Let $\epsilon > 0$ be an (arbitrarily small) real number.
• It follows from the existence of natural numbers exceeding any positive real numbers that there is a natural number $N(\epsilon)$ such that $N > \frac 1\epsilon.$
• It follows that $\frac 1n=|\frac 1n-0| < \epsilon$ for all $n\ge N(\epsilon).$
• By definition of convergence, it means that the real sequence with $x_n:=\frac 1n$, $n\in\mathbb N,$ $n > 0$ is convergent with $\lim_{n\to\infty}\frac 1n=0.$
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References

Bibliography

1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983