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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

    \[ a_n = \frac{(-1)^{n+1}}{n}. \]

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

    \[ \lim_{n \to \infty} \frac{(-1)^{n+1}}{n} \leq \lim_{n \to \infty} \left| \frac{(-1)^{n+1}}{n} \right| = \lim_{n \to \infty} \frac{1}{n} = 0 \quad \implies \quad L= 0. \]

So then we have,

    \begin{align*}  |a_n - L| < \varepsilon && \implies && \left| \frac{(-1)^{n+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.

One comment

  1. tom says:

    I find myself nitpicking here: When you passed to the absolute value you must have assumed the limit L is zero, which is obvious in this case. But beyond knowing the limit in advance Apostol didn’t give any tools for actually determining convergence. What little I know of analysis it seems something like all convergent subsequences sharing the same limit would work, or perhaps cauchy criterion? It just seems that Apostols exposition of sequences was less then comforting; but maybe it will be covered later.

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