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Let $(a_{n})_{n \geq1}$ be a real sequence such that $a_{1}=a_{2}=1$ and $\displaystyle a_{n+2}=a_{n+1}+\frac{a_{n}}{3^n}, n\geq 1$.
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 $$a_{n}=\frac{a_{1}+\ldots+a_{n-2}+3}{3} < 2$$ And what remains to prove it is : $$a_{1}+\ldots +a_{n-2} < 3,$$ but from this point I don't know how I have to do. I need a proof without derivatives. Thanks :) |
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We have $$ a_1=a_2=1,\ a_3=a_2+\frac{a_1}{3}=\frac43<2. $$ If we assume that $a_k<2$ for all $k=1,\ldots,n$, with $n \ge 3$, then we have $$ a_{n+1}=a_2+\sum_{k=2}^n(a_{k+1}-a_k)=a_2+\sum_{k=2}^n\frac{a_{k-1}}{3^{k-1}}<1+\sum_{k=2}^n\frac{2}{3^{k-1}}=1+1-\frac{1}{3^{n-1}}<2. $$ |
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Claim: $$a_n < 2-\frac{1}{3^n} \,.$$ $P(1), P(2)$ are easy to check. Inductive step: $$a_{n+1}= a_{n+1}+\frac{a_n}{3^n} \leq 2-\frac{1}{3^{n+1}}- \frac{2-\frac{1}{3^n}}{3^n}=2-\frac{1}{3^{n+1}}- \frac{2}{3^n}+\frac{1}{9^n} $$ If we can prove taht $$2-\frac{1}{3^{n+1}}- \frac{2}{3^n}+\frac{1}{9^n} < 2-\frac{1}{3^{n+2}}$$ we are done. But this is equivalent to $$\frac{1}{3^{n+2}}+\frac{1}{9^n} <\frac{1}{3^{n+1}}+ \frac{2}{3^n}$$ which is obvious. P.S. This is a pretty standard but not well known technique. If $a_n$ is increasing, then $a_n \leq C$ cannot be proven directly by induction, but one might be able to find a decreasing $b_n \geq 0$, and then prove by induction the stronger claim $$a_n < C-b_n \,.$$ The standard well known example of this phenomena is $$1+\frac{1}{2^2}+..+\frac{1}{n^2} <2 $$ vs $$1+\frac{1}{2^2}+..+\frac{1}{n^2} <2 -\frac{1}{n+1}$$ |
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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 $$ n!f^{(n)}(0)= a_{n} $$ |
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