Consider the sequence \(\{a_n\}\) defined by \(a_n = \frac{(-1)^nn}{n+1}\). Determine whether this sequence converges or diverges. If it converges, find its limit.

Definitions of sequences and convergence: here, Definitions of subsequences: here.
For the “show the limit” template: here.

Solution

We claim that \(\{a_n\}\) does not converge. To see this, suppose for the sake of contradiction that it does converge to some \(L\). Pick \(\epsilon = \frac{1}{2}\). Since it converges, then there must be an \(N \in \mathbb{N}\) such that when \(n \geq N\), we must have

$$ \begin{align*} \left| \frac{(-1)^nn}{n+1} - L \right| &< \frac{1}{2} \end{align*} $$

Observe now that when \(n\) is odd, then \(n=2k+1\) for some \(k\), we see that

$$ \begin{align*} a_{2k+1} &= \frac{(-1)(2k+1)}{(2k+1)+1} \\ &= \frac{-2k-1}{2k+2} \\ &= \frac{k(-2-\frac{1}{k})}{k(2+\frac{2}{k})} \\ &= \frac{-2-\frac{1}{k}}{2+\frac{2}{k}} \\ \end{align*} $$

as \(k\) grows without bound, the \(\{a_{2k+1}\}\) converges to \(-1\). On the other hand, when \(n\) is even, then \(n=2k\) for some \(k\). Then

$$ \begin{align*} a_{2k} &= \frac{(1)(2k)}{(2k)+1} \\ &= \frac{2k}{2k+1} \\ &= \frac{2k}{k(2 + \frac{1}{k})} \\ &= \frac{1}{1 + \frac{1}{k}} \\ \end{align*} $$

as \(k\) grows without bound, the sequence \(\{a_{2k}\}\) converges to \(1\). When \(n \geq N\), both sequences must satisfy

$$ \begin{align*} | a_n - L | < \frac{1}{2} \end{align*} $$

For \(a_{2k}\), this is

$$ \begin{align*} | 1 - L | < \frac{1}{2} \ \longrightarrow \ L \in (1-\frac{1}{2}, 1+\frac{1}{2}) = (0.5,1.5) \end{align*} $$

and for \(a_{2k+1}\), this is

$$ \begin{align*} | -1 - L | < \frac{1}{2} \ \longrightarrow \ L \in (-1-\frac{1}{2}, -1+\frac{1}{2}) = (-1.5, -0.5) \end{align*} $$

However, from this we see that no \(L\) can satisfy both. This is a contradiction. Therefore, the sequence doesn’t converge. \(\blacksquare\)

References

  • Problem Statement Source: Aleph 0