Sequences pop everywhere in math, especially in the world analysis, and they’re closely related to limiting processes. For example, the sum of an infinite series such as \(\sum_{n=0}^{\infty} \frac{1}{n!}\) is really just a sequence of its partial sums:

$$\frac{1}{0!}, \frac{1}{0!} + \frac{1}{1!}, \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!}, \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!}, \cdots$$

And the limit \(\lim_{n \to \infty} \frac{1}{n}\)is also a sequence in the sense that we are just looking for what happens as \(\\n\) grows larger. Below, we look at the values where \(n\) is an increasing power of 10:

$$\frac{1}{1}, \frac{1}{10}, \frac{1}{100}, \frac{1}{1000}, \cdots$$

In this blog post we’re going to take a closer look at these whimsical beings!

Definition

An infinite sequence is just a function \(f: \mathbb N \to X\) from the natural numbers to some other set \(X\). But today we’re only going to be focusing on the case where \(X\) is the set of real numbers! Let’s look at some examples:

• The function \(f: \mathbb N \to \mathbb R\) defined by \(f(n) = 1\) for all \(n \in \mathbb N\) is an infinite sequence, but not an interesting one.

• The function \(g: \mathbb N \to \mathbb R\) given by \(g(n) = \sin n\) is a slightly more interesting sequence, but it doesn’t converge.

• The map \(h: \mathbb N \to \mathbb R\) defined by \(h(n) = \frac{1}{n^2}\) is an infinite sequence that actually converges to a number, namely 0.

This definition is very useful because now we can define the sum of an infinite series (not to be confused with infinite sequences) as the limit of the sequence as \(n \to \infty\) given by \(f(n) = \sum_{k=0}^n \frac{1}{k!}\).

Now that we have a good idea of what a sequence is, let’s look at convergence.

Convergence

In Calculus II you were asked to determine whether a given sequence or series converges using tests. Intuitively, we know what that means: the sequence or series approaches a particular value as we take larger and larger values of \(n\). For example, \(\sum_{n=0}^\infty \frac{1}{n!}\)famously converges to the constant \(e\), or the sequence \((\frac{n}{n+1})_{n=1}^\infty\) converges to \(1\). From here on we’ll be working with the second example, since the former is usually proven using a Taylor series.

Below, we have a very fancy graph of the sequence \(f: \mathbb N \to \mathbb R\) given by \(f(n) = \frac{n}{n+1}\).

As you can see, the curve (if you can call it that) of \(f\) is not a continuous line but rather a collection of points, similar to a scatter plot. This is because the domain of \(f\) is the natural numbers rather than the reals. Notice that when a sequence converges, we can always get as close to it as we want. Suppose that we want \(f\) to be within \(\varepsilon\) units of \(1\). Then we have to show that eventually \(f\) will be within \(\varepsilon\) units of \(1\)and never go outside of this range, because otherwise it’ll be impossible to prove that \(f\) really converges to \(1\). So we want to find an \(N \geq 0\) such that if \(n \geq N\), then \(|f(n) - 1| < \varepsilon\).

Now we can generalize this using our intuition and derive the following definition:

If \((a_n)_{n = 0}^\infty\) is a sequence, we write \(\lim_{n \to \infty} a_n = L\) if and only if for all \(\varepsilon > 0\)there exists some \(N \geq 0\) such that \(|a_n - L| < \varepsilon\)for all \(n \geq N\).

Okay, now that we know what a sequence is and what it really means to converge, let’s go through some examples!

Proving convergence

Let’s say that we want to prove that \((\frac{n}{n+1})_{n=0}^\infty\) converges to \(1\). Typically the way we do these kinds of proofs is to find an \(N \geq 0\) in our scratch work, then prove that the \(N\) we found satisfies the definition above.

Let’s assume that the sequence already converges; our goal is to find an \(N \geq 0\) that satisfies the definition. From the definition, we have

$$\left|\frac{N}{N+1} - 1\right| < \varepsilon$$

We can rewrite the expression inside as \(-\frac{1}{N+1}\) so we are left with \(|-\frac{1}{N+1}| = \frac{1}{N+1} < \varepsilon\). Solving for \(N\), we get

$$\frac{1}{\varepsilon} - 1 < N$$

And for extra safety (mostly to avoid edge cases) we write a similar expression but with a \(\leq\) sign:

$$\frac{1}{\varepsilon} \leq N$$

Perfect! Now in our proof we can use this and show that it really works.

The proof

Let \(\varepsilon > 0\) and let \(N \geq \frac{1}{\varepsilon}\). We claim that \(|\frac{n}{n+1} - 1| < \varepsilon\) for all \(n \geq N\). Indeed, we have \(n \geq \frac{1}{\varepsilon}\), so it follows that \(\frac{1}{\varepsilon} -1 < n\), or \(\frac{1}{\varepsilon} < n + 1.\) Some more algebraic manipulation reveals that

$$\left|-\frac{1}{n+1}\right| = \frac{1}{n+1} < \varepsilon.$$

But \(-\frac{1}{n+1} = \frac{n}{n+1} - 1\) so

$$\left| \frac{n}{n+1} - 1 \right| = |a_n - 1 | < \varepsilon,$$

as needed. Thus we have shown that for all \(\varepsilon > 0\) there exists some \(N \geq 0\) such that \(|a_n - 1| < \varepsilon\) for all \(n \geq N\). Therefore, \((a_n)_{n=0}^\infty\)converges to \(1\).

Conclusion

And that wraps things up! We formalized the definitions of a sequence and the notion of convergence, and even proved that a function converged to a value.