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Find an N such that the convergent sequence is within ε of its limit

Consider the convergent sequence \{ a_n \} with terms defined by

    \[ a_n = \frac{1}{n}. \]

Let L = \lim_{n \to \infty} a_n. Find the value of L and values of N such that |a_n - L| < \varepsilon for all n \geq N for each of the following values of \varepsilon:

  1. \varepsilon = 1,
  2. \varepsilon = 0.1,
  3. \varepsilon = 0.01,
  4. \varepsilon = 0.001,
  5. \varepsilon = 0.0001.

First, we know

    \[ \lim_{n \to \infty} \frac{1}{n} = 0 \quad \implies \quad L= 0. \]

So then we have,

    \begin{align*}  |a_n - L| < \varepsilon && \implies && \left| \frac{1}{n} \right| &< \varepsilon \\  && \implies && \frac{1}{n} &< \varepsilon \\  && \implies && n  &> \frac{1}{\varepsilon}. \end{align*}

Thus, if N > \frac{1}{\varepsilon} then for every n \geq N we have |a_n| < \varepsilon. We compute for the given values of \varepsilon as follows:

  1. \varepsilon = 1 implies N = \frac{1}{1} = 1.
  2. \varepsilon = 0.1 implies N = \frac{1}{.1} = 10.
  3. \varepsilon = 0.01 implies N = \frac{1}{.01} = 100.
  4. \varepsilon = 0.001 implies N = \frac{1}{.001} = 1000.
  5. \varepsilon = 0.0001 implies N = \frac{1}{.0001} = 10000.

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