Let \(\{a_n\}\) be a sequence. Prove that if \(\lim\limits_{n \rightarrow \infty} a_n = L\) and \(\lim\limits_{n \rightarrow \infty} a_{n+1} = M\), then \(L = M\).

For the definitions of sequences and what it means to for a sequence to converge, see this, for subsequences see this.
For the “show the limit” template, see this.

Formal Proof

Since \(\lim a_n = L\), then by definition this means that there exists some \(N_1 \in \mathbb{N}\) such that for all \(n \geq N_1\), we must have

$$ \begin{align*} \big\lvert a_n - L \big\rvert &< \epsilon \end{align*} $$

Similarly there exists some \(N_2 \in \mathbb{N}\) such that for all \(n \geq N_2\), we must have

$$ \begin{align*} \big\lvert a_{n+1} - M \big\rvert &< \epsilon \end{align*} $$

Let \(N = \max(N_1,N_2)\). Then for all \(n \geq N\), we have

$$ \begin{align*} \big\lvert a_n - L \big\rvert &< \epsilon \\ \big\lvert a_{n+1} - M \big\rvert &< \epsilon \end{align*} $$

Let \(m = n + 1\). Then, whenever \(m \geq N + 1\),

$$ \begin{align*} \big\lvert a_m - L \big\rvert &< \epsilon \\ \big\lvert a_m - M \big\rvert &< \epsilon \end{align*} $$

Observe now that

$$ \begin{align*} \big\lvert L - M \big\rvert &= \big\lvert (a_m - L) - (a_m - M) \big\rvert \end{align*} $$

We can now apply the triangle inequality along with the observation that for any \(\epsilon > 0\), since \(a_m \rightarrow L\) and \(a_m \rightarrow M\), there exists an \(N\) such that for \(m \geq N\)

$$ \begin{align*} \big\lvert L - M \big\rvert = \big\lvert (a_m - L) - (a_m - M) \big\rvert &\leq \big\lvert a_m - L\big\rvert + \big\lvert a_m - M \big\rvert \\ &< \epsilon + \epsilon = 2\epsilon \\ \end{align*} $$

Since \(\epsilon\) was arbitrary, this implies that \(\big\lvert L - M \big\rvert = 0\), and hence \(L = M\) as we wanted to show. \(\blacksquare\)

References

  • Problem Statement Source: Aleph 0