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Let $a_0 = 0$, $a_1 = 0$, $a_2=1$ and for $n>2$, $a_n = \dfrac{a_{n-1}+a_{n-2}+a_{n-3}}{3}$. Consider $\lim\limits_{n \to +\infty} a_n$.

Using a python script I found that $a_n$ tends to $\frac{1}{2}$ as $n \to +\infty$. However I don't know how to prove this...I tried the approach from this question that I asked here a while back, but I couldn't anything like it to work because the differences between terms kept altering between positive and negative in an unpredictable fashion (i.e. -+-++-++-++-+--+--...) and the absolute value of the differences did not always decrease as $n$ increased ($|a_7-a_8| < |a_8-a_9|$). (Source: the python code, which I can include if that helps...)

Another approach I tried was to use induction to try to prove the general case that if the first $k$ terms $a_0,\cdots,a_{k-1}$ are given where $a_0,\cdots,a_{k-2} = 0$ and $a_{k-1} = 1$ and for $n>k-1$,

$a_n = \dfrac{\sum\limits_{i=1}^k a_{n-i}}{k}$

then that $\lim\limits_{n \to +\infty} a_n = \dfrac{2}{k+1}$ (this is just a guess after the first five or six terms), but I also made no progress there.

Could anybody help point me in the right direction? How do I prove this? (The $k=3$ case is enough; proving the general case was just an idea.)

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One way is to solve the recurrence explicitly, by using the characteristic polynomial. – André Nicolas Aug 20 '14 at 14:33
See j.c.'s answer here for a proof and a more general approach! – user161516 Aug 20 '14 at 14:36
up vote 2 down vote accepted

There is a way to solve this problem, but it requires time and work. The recurrence relation for this sequence is similar to that of the Fibonnaci sequence, in general, these types of sequences have closed forms in terms of a sum of exponentials of n.

Here is how you can solve the problem: by the linearity of the recurrence relation, if $s_n$ is a sequence which satisfies the relation, then $ks_n$ also does ( $k$ is any value). If also, say $t_n$ satisfies the relation, then the sum $$s_n+t_n $$ also does. Hence, to find the closed form of the sequence, you must find three distinct sequences which satisfy the recurrence relation. Then, your sequence is of the form $$k_1 s_n + k_2t_n + k_3u_n $$ where $s_n$, $t_n$ and $u_n$ are the 3 sequences you found.

To find the 3 sequences, assume they are of exponential form: $$s_n= x^n $$ for some constant $x$. Substituting in the recurrence relation and simplifying, youre left with:

$$3x³=x²+x+1 $$ I leave you to solve that ( hint, one solution is easy by inspection, then factor the cubic and solve the final quadratic)

You will find that two of the roots are complex and one is real. Then, find your $k_1$, $k_2$, $k_3$ mentioned before by substituting the the values for $n=0,1,2$

Whew. ( I know it's long) Finally, take the limit as $n \to \infty$. You will find two of the exponentials vanish ( hint: what is their modulus?) and the third is trivial. That will yield your limit to be $\frac{1}{2}$.

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The calculation that you have very nicely described works smoothly and quickly. – André Nicolas Aug 20 '14 at 14:45
Thank you Andre Nicolas. It can get quite tedious sometimes. – Assaultous2 Aug 20 '14 at 14:46

Use a telescoping sum to get $$3a_n+2a_{n-1}+a_{n-2}=3a_2+2a_1+a_0.$$ Now it is easy to find the answer, by taking $n\to\infty$ in the above equation: $${3\cdot 1+2\cdot0+1\cdot0\over3+2+1}=\frac12.$$ Actually, we first have to prove that $\lim_{n\to\infty}a_n$ exists.

With this method, you can also prove the general case.

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$\lim_{n\to\infty}(3a_n+2a_{n-1}+a_{n-2})=6a_{\infty}$. Since $3a_2+2a_1+a_0=3+0+0$, we obtain $a_{\infty}=1/2$. Nice results! – mike Aug 20 '14 at 14:38
@mike How do I show the step of taking the limit more rigorously? – user90667 Aug 20 '14 at 14:49
Please take a look at the following MSE question and Answer.… – mike Aug 20 '14 at 14:53

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