How do you compute the first few terms of the recursively defined sequence $a_{n+1}=\frac{1}{3}(a_n+1)$ with $a_1=1$?

Conditions

• The recurrence relation is $a_{n+1}=\frac{1}{3}(a_n+1)$.
• The initial value is $a_1=1$.
• The index $n$ starts at $1$.

Reasoning, step by step

1. Substitute $a_1=1$ into the recurrence for $n=1$: $a_2=\frac{1}{3}(1+1)$.
2. Simplify the arithmetic to get $a_2=\frac{2}{3}$.
3. Substitute $a_2=\frac{2}{3}$ into the recurrence for $n=2$: $a_3=\frac{1}{3}(\frac{2}{3}+1)$.
4. Add the fractions inside the parentheses: $\frac{2}{3}+1=\frac{5}{3}$.
5. Multiply by $\frac{1}{3}$ to get $a_3=\frac{5}{9}$.

Example

The board shows the explicit calculations: $a_2=\frac{1}{3}[1+1]=\frac{2}{3}$ and $a_3=\frac{1}{3}\left[\frac{2}{3}+1\right]=\frac{5}{9}$.

Common misconceptions

• Assuming the recurrence applies to $n=0$ when the domain is specified as $n \ge 1$.
• Forgetting to distribute the $\frac{1}{3}$ to both the $a_n$ and the $1$ in the recurrence.

Watch the explanation

Explore next

• What lower-bound claim is made about the sequence terms?
• How is the statement $a_n \ge \frac{1}{2}$ proved by induction?