Proving that a series is bounded from above through induction.

Question

I am required to prove that the following series $$a_1=0, a_{n+1}=(a_n+1)/3, n \in N$$ is bounded from above and is monotonously increasing through induction and calculate its limit. Proving that it's monotonously increasing was simple enough, but I don't quite understand how I can prove that it's bounded from above through induction, although I can see that it is bounded. It's a fairly new topic to me. I would appreciate any help on this.

Answer 1

See that for any $a_n<\frac12$, we have

$$a_{n+1}=\frac{a_n+1}3<\frac{\frac12+1}3=\frac12$$

Thus, it is proven that since $a_0<\frac12$, then $a_1<\frac12$, etc. with induction.

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We choose $\frac12$ since, when solving $a_{n+1}=a_n$, we result with $a_n=\frac12$, the limit of our sequence.

Answer 2

Hint

Let $f(x)=\frac{x+1}{3}$

$f$ has one fixed point $L=\frac{1}{2}=f(L)$

Now, you can prove by induction that

$$\forall n\geq 0\;\;a_n\leq L$$

using the fact that $f$ is increasing at $\; \mathbb R$ and $a_{n+1}=f(a_n)$.

Answer 3

We have

$$\begin{cases}a_0&=0\\a_1&=\frac13 \\a_2& =\frac 13+\frac{1}{3^2}\\a_3& =\frac 13+\frac{1}{3^2}+\frac{1}{3^3} \end{cases}$$ What does this sugest? That $$a_n=\sum_{k=1}^n\frac{1}{3^k}=\dfrac{\frac13-\frac{1}{3^{n+1}}}{\frac23}.$$ Show this by induction and get that $a_n<\frac 12,\forall n\in \mathbb{N}.$