# Accumulation Points of a Sequence

Definition: Let $(a_n)$ be a sequence of real numbers. The number $a$ is said to be an accumulation point of $(a_n)$ if there exists a subsequence $(a_{n_k})$ such that $\lim_{k \to \infty} a_{n_k} = a$, that is, $\forall \epsilon > 0$ $\exists K \in \mathbb{N}$ such that if $k ≥ K$ then $\mid a_{n_k} - a \mid < \epsilon$. |

For example, consider the sequence $\left ( \frac{1}{n} \right )$ which we verified earlier converges to $0$ since $\lim_{n \to \infty} \frac{1}{n} = 0$. If we take the subsequence $(a_{n_k})$ to simply be the entire sequence, then we have that $0$ is an accumulation point for $\left ( \frac{1}{n} \right )$.

As another example, consider the sequence $((-1)^n) = (-1, 1, -1, 1, -1, ... )$. This sequence does not converge, however, if we look at the subsequence of even terms we have that it's limit is 1, and so $1$ is an accumulation point of the sequence $((-1)^n)$. If we look at the subsequence of odd terms we have that its limit is -1, and so $-1$ is also an accumulation point to the sequence $((-1)^n)$.

Theorem 1: Let $(a_n)$ be a sequence of real numbers. If $(a_n)$ converges to $L$, then $L$ is the only accumulation point of $(a_n)$. |

**Proof:**Suppose that $(a_n)$ is a sequence of real numbers that converges to $L$. Then by one of the theorems on the Subsequences page, all convergent subsequences $(a_{n_k})$ converge to $L$, and so $L$ is the only accumulation point of $(a_n)$. $\blacksquare$

Theorem 2: Let $(a_n)$ be a bounded sequence of real numbers. Then $(a_n)$ has at least one accumulation point. |

**Proof:**Since $(a_n)$ is bounded, by The Bolzano-Weierstrass Theorem, $(a_n)$ contains a convergent subsequence $(a_{n_k})$. Suppose that $(a_{n_k})$ converges to $L$. Then $L$ is an accumulation point for $(a_n)$. $\blacksquare$

As a remark, we should note that theorem 2 partially reinforces theorem 1. Recall that a convergent sequence of real numbers is bounded, and so by theorem 2, this sequence should also contain at least one accumulation point. Theorem 1 however, shows that provided $(a_n)$ is convergent, then this accumulation point is unique.

Theorem 3: If $(a_n)$ is a properly divergent subsequence then $(a_n)$ has no accumulation points. |

**Proof:**We saw from the Properly Divergent Sequences page that if a sequence $(a_n)$ is properly divergent then any subsequence $(a_{n_k})$ is not convergent, and so there exists no accumulation points for $(a_n)$. $\blacksquare$

Now let's look at some examples of accumulation points of sequences.

## Example 1

**Let $(a_n)$ be a sequence defined by $a_n = \left\{\begin{matrix} 1/n & \mathrm{if \: n = 2k} \\ n & \mathrm{if \: n =2k - 1} \end{matrix}\right.$. Show that there exists only one accumulation point for $(a_n)$.**

If we look at the sequence of even terms, notice that $\lim_{k \to \infty} a_{2k} = 0$, and so $0$ is an accumulation point for $(a_n)$. Now let's look at the sequence of odd terms, that is $\lim_{k \to \infty} a_{2k-1} = \lim_{n \to \infty} n = \infty$. Since the terms of this subsequence are increasing and this subsequence is unbounded, there are no accumulation points associated with this subsequence and there are no accumulation points associated with any subsequence that at least partially depends on the tail of this subsequence.

We deduce that $0$ is the only accumulation point of $(a_n)$.

## Example 2

**Let $(a_n)$ be a sequence defined by $a_n = \frac{n + 1}{n}$. Determine all of the accumulation points for $(a_n)$.**

Notice that $a_n = \frac{n+1}{n} = 1 + \frac{1}{n}$. We know that $\lim_{n \to \infty} 1 + \frac{1}{n} = 1$, and so $(a_n)$ is a convergent sequence. By theorem 1, we have that all subsequences of $(a_n)$ must therefore converge to $1$, and so $1$ is the only accumulation point of $(a_n)$.

## Example 3

**Consider the sequence $(a_n)$ defined by $a_n = \left\{\begin{matrix} n & \mathrm{if \: 6 \: divides \: n }\\ n^2 & \mathrm{if \: 6 \: does \: not \: divide \: n} \end{matrix}\right.$. Does $(a_n)$ have accumulation points?**

Notice that $(a_n)$ is constructed from two properly divergent subsequences (both that tend to infinity) and in fact $(a_n)$ is a properly divergent sequence itself. Therefore, there does not exist any convergent subsequences, and so $(a_n)$ has no accumulation points.