How to prove $a_{n} < 2$ if $\displaystyle a_{n+2}=a_{n+1}+\frac{a_{n}}{3^n}$ Let $(a_{n})_{n \geq1}$ be a real sequence such that $a_{1}=a_{2}=1$ and $\displaystyle a_{n+2}=a_{n+1}+\frac{a_{n}}{3^n}, n\geq 1$. 


  Prove that $a_{n} &lt; 2, \forall n \geq 1.$


I write $$\sum a_{k+2}-a_{k+1}=\sum \frac{a_{k}}{3}$$ 
and I obtained : 

$$3a_{n+2}=a_{1}+\ldots+a_{n}+3$$

or

$$a_{n}=\frac{a_{1}+\ldots+a_{n-2}+3}{3} &lt; 2$$

And what remains to prove it is : 

$$a_{1}+\ldots +a_{n-2} &lt; 3,$$ but from this point I don't know how I have to do. 

I need a proof without derivatives. 

Thanks :)
We have
$$
a_1=a_2=1,\ a_3=a_2+\frac{a_1}{3}=\frac43&lt;2.
$$
If we assume that $a_k&lt;2$ for all $k=1,\ldots,n$, with $n \ge 3$, then we have 
$$
a_{n+1}=a_2+\sum_{k=2}^n(a_{k+1}-a_k)=a_2+\sum_{k=2}^n\frac{a_{k-1}}{3^{k-1}}&lt;1+\sum_{k=2}^n\frac{2}{3^{k-1}}=1+1-\frac{1}{3^{n-1}}&lt;2.
$$
if we assume there is a generating function so

$$ f(x)= \sum_{n=0}^{\infty}a_{n}x^{n} $$

then $ f(x) $ satisfy the functional equation

$$ f(x)-a_{0}-a_{1}x=f(x)x-a_{0}x+x^{2}f(x/3) $$

from this i think you could obtaien the derivatives so $$ n!f^{(n)}(0)= a_{n} $$