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Assume that $$\lim_{n\to\infty}(a_{n+2}-a_{n})=A$$ show that

$\displaystyle \lim_{n\to\infty}\dfrac{a_{n}}{n}$and $\displaystyle\lim_{n\to\infty}\dfrac{a_{n+1}-a_{n}}{n}$ exist and find these limits.

maybe this problem have some methods, Thank you.

since $$\lim_{n\to\infty}(a_{n+2}-a_{n})=A$$ so there exists $N$, and for $\forall \varepsilon>0$, such $$|a_{n+2}-a_{n}-A|\le\varepsilon$$

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It is given that for every $\varepsilon>0$ there exist a $N_0$ such that $n>N_0\implies$ $|a_{n+2}-a_n-A|<\varepsilon\implies a_n+A-\varepsilon<a_{n+2}<a_n+A+\varepsilon $

$a_{N'}+\frac{(n-2-N')}{2}A-\frac{(n-2-N')}{2}\varepsilon <a_{n}<a_{N'}+\frac{(n-2-N')} {2}A+\frac{(n-2-N')}{2}\varepsilon \implies\\ \frac{a_{N'}}{n}+\frac{A}{2}-\frac{(2+N')}{2n}\cdot A-\varepsilon+\frac{(2+N')}{2n}\varepsilon<\frac{a_n}{n}<\frac{a_{N'}}{n}+\frac{A}{2}-\frac{(2+N')}{2n}\cdot A+\varepsilon-\frac{(2+N')}{2n}\varepsilon \implies \\ \boxed{\lim\limits_{n\rightarrow \infty}{\frac{a_n}{n}}=\frac{A}{2}}$

Where $N'$ is $N_0$ when both $n$ and $N_0$ have same parity and $N_0+1$ otherwise.

Similarly, forming upper and lower bounds for $a_n-a_{n-1}$ we can arrive at $\lim\limits_{n\rightarrow\infty}{\frac{a_n-a_{n-1}}{n}}=0$

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If $(c_N)_{N\geqslant 1}$ is a sequence of real numbers converging to $l$, then $N^{-1}\sum_{j=1}^Nc_j\to l$. With $c_N=a_{2N}$, we have $c_{n+1}-c_n\to 0$, hence $\frac{a_{2n}}{2n}\to A/2$. With $c_N=a_{2N+1}$, we obtain that $\frac{a_{2n+1}}{2n+1}\to A/2$.

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  • $\begingroup$ What about the first claim, how do you justify it? $\endgroup$ – chubakueno May 18 '14 at 15:21
  • $\begingroup$ @chubakueno It is Cesaro convergence: use the definition of the limit with $\varepsilon$, and split the sum where the terms are close up to $\varepsilon$ to the limit. The contribution of the others is neglectible. $\endgroup$ – Davide Giraudo May 18 '14 at 15:25
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Hint: use the identities of the kind $$ d_n= a_{n+1} -a_{n} = a_{n+1}-a_{n-1}+a_{n-2} -a_{n} +a_{n-1}-a_{n-2} = a_{n+1}-a_{n-1}+a_{n-2} -a_{n} + d_{n-2} $$

Answers: $\lim\limits_{n\rightarrow +\infty} a_n/n = A/2$ and $\lim\limits_{n\rightarrow +\infty} (a_{n+1}-a_n)/n=0$

Example as a check: take $a_n=nA/2$.

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  • $\begingroup$ No,maybe this methods can't works. $\endgroup$ – china math May 18 '14 at 15:21

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