How do you compute $a_{n+1}-a_n$ for the recurrence $a_{n+1}=\frac{1}{3}(a_n+1)$?

Conditions

• The recurrence relation is $a_{n+1}=\frac{1}{3}(a_n+1)$.
• The goal is to analyze the sign of the difference to determine monotonicity.

Reasoning, step by step

1. Start with the recurrence: $a_{n+1}=\frac{1}{3}(a_n+1)$.
2. Subtract $a_n$ from both sides: $a_{n+1}-a_n = \frac{1}{3}(a_n+1) - a_n$.
3. Distribute the $\frac{1}{3}$: $a_{n+1}-a_n = \frac{1}{3}a_n + \frac{1}{3} - a_n$.
4. Combine the $a_n$ terms: $\frac{1}{3}a_n - a_n = -\frac{2}{3}a_n$.
5. Write the final simplified difference: $a_{n+1}-a_n = -\frac{2}{3}a_n + \frac{1}{3}$.

Example

The lecturer writes $a_{n+1}-a_n=\frac{1}{3}a_n+\frac{1}{3}-a_n=-\frac{2}{3}a_n+\frac{1}{3}$.

Common misconceptions

• Incorrectly distributing the $\frac{1}{3}$ only to $a_n$ and not to the constant $1$.
• Failing to combine $\frac{1}{3}a_n$ and $-a_n$ correctly.

Watch the explanation

Explore next

• Why does proving $-\frac{2}{3}a_n+\frac{1}{3}\le0$ reduce to the already known bound $a_n\ge\frac{1}{2}$?
• Does this example prove the sequence is non-decreasing or non-increasing?