Limits of sequences

Basic idea

Sequences are also functions, and as a result many of the concepts we introduced for functions apply to sequences as well. For instance, we may:

  • make a table out of them

  • plot them in some relevant window

  • describe in detail some slices of it; in particular the region when \(n\) is large.

In these notes we will focus on the later.

Consider the sequence:

\[ u_n=2+\frac{1}{n} \]whose graph for \(n=1,.. ,20\) is:

A few things strikes us in the region when \(n\) is large:

  1. As \(n\) gets larger and larger, then \(2+1/n\) gets closer and closer to \(2\).
  2. The values \(u_n\) never actually reaches the value of \(2\) because \(1/n\) is never zero, no matter how large \(n\) is.

In other word: the sequence \(u_n\) eventually lies within any interval centered at \(2\).

Lets make this wording rigorous:

  • To say that an interval centered at \(2\) contains a sequence means: \(|2-u_n|<\varepsilon\). The radius of the interval is \(\varepsilon\) and is positive, because we are speaking about an interval, not a point like thing!

  • The eventually words means that after some value \(N\), any \(n\), guarantees \(|2-u_n|<\varepsilon\)

  • The any word means, that the above is true for any positive \(\varepsilon\)

Putting it all together:

  • (1) For any \(\varepsilon>0\), the following is true:

    • (2) That there exists some \(N\) that guarantees that:

      • (3) for any \(n\geq N\) the following statement is true

        • (4) the statement that \(|2-u_n|<\varepsilon\)

We write this description using complicated mathematical notation as:

\[ \overbrace{\forall \varepsilon >0 }^{(1)}:\overbrace{\exists N}^{(2)}:\overbrace{\forall n\geq N}^{(3)}: (\overbrace{ |2-u_n|<\varepsilon)}^{(4)} \tag{1}\]

Since it is just complicated to write Equation 1, the following abbreviation is introduced:

\[ \lim_{n\longrightarrow \infty}u_n = 2 \]

The lhs should be view as a single symbol that says: “the limit value of \(2+1/n\) when \(n\) gets larger and larger (infinity is not a number)”. The rhs tell us that \(2\) is the limit value; or better said, the point around which the sequence eventually always lurks.

Commentary

Sequences of the form \(x_0\pm1/n\) are important since they describe a step by step approach to the number \(x_0\) from values above \(+\) or below \(-\); without never actually getting there. They will be relevant when computing.

Proving a sequence converges

An example

We will prove that:

\[ \lim \frac{4n+1}{2n}=2 \]

Meaning we want to prove that:

\[ \forall \varepsilon >0:\exists N:\forall n\geq N : |2-(4n+1)/2n|<\varepsilon \]

We proceed by unpacking the meaning with indented assumptions:

  • Assume we have some value \(\varepsilon\), though it is arbitrary:

    • We seek to find an \(N\) that makes \(\forall n\geq N : |2-(4n+1)/2n|<\varepsilon\) true, i.e., we seek a lower bound on \(n\) that makes this true:

      • To do that, we assume that we have some value \(n^*\) which is larger than the \(N\) and which makes the statement \(|2-(4n^*+1)/2n^*|<\varepsilon\) true. This much we know about \(n^*\). No special conclusion was taken, yet.

        Now simplify this statement and see if we find something new about \(n^*\):

        \[ \begin{align} & |2-\frac{4n^*+1}{2n^*}|<\varepsilon\\ &\implies |\frac{4n^*-4n^*+1}{2n^*}|<\varepsilon\\ &\implies |\frac{1}{2n^*}|<\varepsilon\\ &\implies n^*>\frac{1}{2\varepsilon} \end{align} \]

Now, this is significant! Lets recollect what we found in the three indentations: Given a radius \(\varepsilon\), a \(n^*\)bounded below by \(N\) and which obeys our statement is a \(n^*\) larger than \(1/2\varepsilon\). The consequences of this are inescapable. If you choose smaller and smaller \(\varepsilon\) the larger and larger \(n^*\) must be. \(\varepsilon\) can be make as small as we wish and we can always find, through the inequality \(n^*>1/2\varepsilon\), a \(n^*\) that makes \(|2-(4n^*+1)/2n^*|<\varepsilon\) true. And since \(N\) is smaller than \(n^*\) we can, in conclusion, always say it exist a lower bound \(N\) for this \(n^*\).

To say \(\lim \frac{4n+1}{2n}=2\) is just an abbreviation for the whole reasoning made above!

Basic sequences limits

Using arguments such as the one shown above we can establish a myriad of limits for basic sequences, we are going ino the details of the proofs and just list the results:

  • \(\lim \frac{a}{b}=\frac{a}{b}\) provided \(b\not = 0\)

  • \(\lim \frac{an+b}{cn+d}=\frac{a}{c}\) provided \(c\not = 0\)

  • \(\lim \frac{1}{n^p} =0\), if \(p>0\)

  • so on …

We can also prove basic rules for limits, an important one is:

  • If \(a_n\) and \(b_n\) are convergent to limits \(a\) and \(b\), then:

    • the sequence \(a_n+b_n\) converges to \(a+b\).

    • the sequence \(a_nb_n\) converges to \(ab\)

    • the sequence \(a_n/b_n\) converges to \(a/b\) if \(b\not=0\)

    • so on…

The idea behind these pre-made limits and the rules is similar to the one from derivatives: If you want to compute a limit of complicated sequence, then your goal is to rearrange that sequence and the limit in such a way that the basic pre-made limits can be used. Giving us the limit value of the complicated sequence.