More on Sequences — Question 6

Question 6 – Solution

Step 1: Prove convergence.

Use absolute values to remove the alternating sign: $$|a_n-0|=\frac1n\longrightarrow0,$$ Since $1/n\to0$, the definition of convergence gives $a_n\to0$. Equivalently, $-1/n\le a_n\le1/n$ and the Squeeze Theorem applies.

Step 2: Test eventual monotonicity.

For every positive integer $k$, $$a_{2k}=\frac1{2k}>0,\qquad a_{2k+1}=-\frac1{2k+1}<0.$$ Thus $a_{2k}>a_{2k+1}$ but $a_{2k+1}<a_{2k+2}$, producing infinitely many decreases and increases. Because this occurs for arbitrarily large $k$, no tail can be monotone.

Step 3: Analyze the two subsequences.

The even subsequence $a_{2k}=1/(2k)$ decreases to $0$, while the odd subsequence $a_{2k-1}=-1/(2k-1)$ increases to $0$. Both close in on the same limit from opposite sides. This demonstrates that convergence does not require eventual monotonicity.