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

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 (since 0 < \frac{1}{n!} \leq \frac{1}{n} for all n) that

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

So then we have,

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

Thus, if N > \frac{1}{\varepsilon} then N! > \frac{1}{\varepsilon} as well (since N! > N). So, 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.

2 comments

    • RoRi says:

      I don’t understand where you get those values of $N$. I think the above is correct, but maybe I have made a mistake somewhere?

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):