Find the limit of $d_1 = 1, d_{n+1}=1+\frac{d_n}{3}$

Question

Find the limit of $d_1 = 1, d_{n+1}=1+\frac{d_n}{3}$

We show $d_n\leq(3/2)$ by induction.

$d_1=1 = (2/2)\leq(3/2)$ and so our hypothesis is that $d_n\leq(3/2)$.

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$n+1:$

$d_{n+1} \leq 1 + \frac{(3/2)}{3} = (3/2)$

Now, if we assume the limit $L$ exists, it must satisfy the equation $L=1+(L/3)$, which is satisfied by $L=(3/2)$.

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Can I already conclude that thus our equation is bounded by $1\leq d_n\leq (3/2)$ and converges towards $(3/2)$? Or do I have to show the sequence is increasing first through induction as well probably (which is quickly done tbf), or is this implicitly contained in the induction above?

Answer 1

$|d_{n+1}-d_n|=|1+{d_n\over 3}-(1+{d_{n-1}\over 3})|={1\over 3}|d_n-d_{n-1}|$. Deduce that $|d_{n+1}-d_n|\leq {1\over 3^n}|d_1-d_0|$ and the sequence is a Cauchy sequence.

Answer 2

The equation $L=1+\frac{L}{3}$ has one solution, which is $L=\frac{3}{2}$. Since $\{d_n\}_{n \to \infty}$ is increasing and bounded (as you noted, by $\frac{3}{2}$), it has a limit. Thus, if $L:=\lim_{n \to \infty} d_n,$ then $L=\frac{3}{2}$.

Answer 3

First you need to show that the sequence is increasing and bounded above, thus it is convergent.

Then you apply the equation $L=1+L/3 $ to find $L=3/2$