Prove that for all k>100, the quantity (1)/(k^2) is in the...

Prove that for all $k>100,$ the quantity $\frac{1}{k^{2}}$ is in the interval (0,0.0001) . What does this have to do with the limit of the sequence $\left\{\frac{1}{k^{2}}\right\}$ as $k \rightarrow \infty$ ?

Key Concepts

Bounding with Inequalities

Bounding a sequence term using inequalities is a common technique in analysis. It involves showing that the terms of a sequence are less than a given positive number once the index exceeds a particular value. This method is directly applied in the question by proving that 1/k² remains below 0.0001 for all k sufficiently large.

Convergence and Tail Behavior

Convergence of a sequence is closely related to the behavior of its tail—the part of the sequence beyond some index. If the tail of the sequence lies entirely within a small interval around the limit, the sequence is converging toward that limit. In this case, showing that every term for k > 100 falls within the given bound helps underline that the sequence converges to 0.

Epsilon Definition of Limits

This definition formalizes the idea of limits by stating that for every ? > 0, there exists an index N such that for all k > N, the absolute difference between the sequence term and the limit is less than ?. In this problem, 0.0001 plays the role of ?, demonstrating that beyond a certain index, the terms are confined within an ?-interval around the limit.

Limit of a Sequence

A limit of a sequence is a value that the terms of the sequence approach as the index goes to infinity. The concept is fundamental in analysis and is used to define convergence, where a sequence is said to converge to a limit if for every specified degree of closeness, the sequence's terms eventually remain that close to the limit.

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