Sequences — Question 5

*Question 5 Let $a_n=\sin(n\pi/2)$ for $n\ge1$.

1. Make a table giving $a_n$ for each residue class of $n$ modulo $4$, and list the first eight terms.

2. Determine whether the sequence converges. Justify your answer using subsequences.

3. Explain whether changing or rearranging finitely many initial terms can change convergence.

*Question 5 – Solution

Since adding $2\pi$ does not change sine and increasing $n$ by $4$ adds $2\pi$ to $n\pi/2$, the values repeat every four indices: $$\begin{array}{c|cccc} n\pmod4&1&2&3&0\\ \hline a_n&1&0&-1&0 \end{array}$$ Thus the first eight terms are $1,0,-1,0,1,0,-1,0$.

In particular, $$a_{4k+1}=1\quad\text{and}\quad a_{4k+3}=-1$$ for all applicable $k$. The first subsequence converges to $1$, while the second converges to $-1$.

If $(a_n)$ converged, all its subsequences would share the same limit. These constant subsequences have different limits, so $(a_n)$ diverges.

Changing, deleting, or rearranging only finitely many initial terms cannot change convergence because the definition concerns all terms beyond some index. No finite modification removes the infinitely recurring values $1$ and $-1$. Even an arbitrary permutation of all these terms still contains infinitely many occurrences of both values, so it cannot converge.