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 $$|a_{n+2}-a_{n}-A|\le\varepsilon$$


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 $\lim\limits_{n\rightarrow\infty}{\frac{a_n-a_{n-1}}{n}}=0$


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 $\frac{a_{2n+1}}{2n+1}\to A/2$.

  • $\begingroup$ 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

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 $a_n=nA/2$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.