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

by RoRi

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

    \[ a_n = \frac{2n}{n^3 + 1}. \]

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

    \[ \lim_{n \to \infty} \frac{2n}{n^3 + 1} = \lim_{n \to \infty} \frac{\frac{2}{n^2}}{1 + \frac{1}{n^3}} = 0 \quad \implies \quad L= 0. \]

So then we have,

    \begin{align*} |a_n - L| < \varepsilon && \implies && \left| \frac{2n}{n^3+1} \right| &< \varepsilon \\[9pt] && \implies && \frac{2n}{n^3 + 1} &< \varepsilon \\[9pt] && \implies && 2n &< \varepsilon (n^3+1) \\[9pt] && \implies && \frac{2}{\varepsilon} &< n^2 + \frac{1}{n}. \end{align*}

Thus, if N > \sqrt{\frac{2}{\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 \geq \sqrt{\frac{2}{1}} = \sqrt{2}.
  2. \varepsilon = 0.1 implies N \geq \sqrt{\frac{2}{.1}} = \sqrt{20} = 2 \sqrt{5}.
  3. \varepsilon = 0.01 implies N \geq \sqrt{\frac{2}{.01}} = \sqrt{200} = 10\sqrt{2}.
  4. \varepsilon = 0.001 implies N \geq \sqrt{\frac{2}{.001}} = \sqrt{2000} = 20 \sqrt{5}.
  5. \varepsilon = 0.0001 implies N \geq \sqrt{\frac{2}{.0001}} = \sqrt{20000} = 100 \sqrt{2}.

6 comments

      • Mohammad Azad says:

        This blog’s comment system is so bugged. I will try to use words instead, remove the one from the denominator, then cancel the ns and get reduse the power from 2 to 1, then n greater than two over epsilon works.

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