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What is the right way to solve a problem like this?

Let $x_n$ a sequence of real numbers such that: $$ \lim_{n \to \infty}(x_n - x_{n+2})=0,$$ So try that: $$ \lim_{n \to \infty}\frac{x_n - x_{n+1}}{n}=0.$$

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It is perhaps interesting that the denominator $n$, which at first sight looks unnecessarily large, cannot be replaced by something substantially smaller, like $\frac{n}{\log n}$. –  André Nicolas Jan 3 '12 at 15:50

3 Answers 3

Let $\epsilon\gt0$ be given. Find $n_0\in\mathbb N$ such that $|x_n-x_{n+2}|\lt\epsilon/2$ for $n\ge n_0$. Then

$$ \begin{eqnarray} \left|\frac{x_{n_0+2k}-x_{n_0+2k+1}}{n_0+2k}\right| &\le& \frac{ \left|x_{n_0}-x_{n_0+1}\right|+ \sum_{j=0}^{k-1}\;\left|x_{n_0+2j}-x_{n_0+2j+2}\right|+ \sum_{j=0}^{k-1}\;\left|x_{n_0+2j+1}-x_{n_0+2j+3}\right| }{n_0+2k} \\ &\le& \frac{ \left|x_{n_0}-x_{n_0+1}\right|+2k\epsilon/2 }{n_0+2k} \;. \end{eqnarray} $$

For large $k$, this tends to $\epsilon/2$ and is thus eventually less than $\epsilon$.

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The result follows by applying the Stolz-Cesàro theorem to the sequence $y_n = (-1)^n \left( x_{n} - x_{n-2} \right)$.

Since $y_n \to 0$, the theorem guarantees that $\frac1N \sum \limits_{n=1}^N y_n \to 0$ as $N \to 0$. Now by a telescopic cancelation taking the odd and even terms separately (explained below), we have $$ \begin{align*} \sum_{n=1}^N y_n &= (-1)^N y_N + (-1)^{N-1}y_{N-1} - (-1)^1 y_1 - (-1)^0 y_0 \\ &= (-1)^{N} \left( y_{N} - y_{N-1} \right) + O(1). \end{align*} $$ From this, it is easy to get the desired claim. $\qquad \diamond$


Intuition. Of course, there is no one right way to solve any problem. I will explain my train of thoughts between the problem statement and the above solution, with the hope that you find it helpful.

If you play with the given condition, it is evident that it is just two separate conditions bundled together into one; i.e., it says that the successive terms of the odd (resp. even) subsequence $\sum x_{2n-1}$ (resp. $\sum x_{2n}$) get closer to each other. More precisely, we have the condition $$ x_{2n+1} - x_{2n-1} \to 0, \tag{$\dagger$} $$ and separately, $$ x_{2n} - x_{2n-2} \to 0. \tag{$\ddagger$} $$

Now, notice that both $(\dagger)$ and $(\ddagger)$ remind us of a telescopic cancelation: $$ \begin{align*} x_{2N+1} &= \sum_{n \leqslant N} \left(x_{2n+1} - x_{2n-1} \right) + x_1 \\ x_{2N} &= \sum_{n \leqslant N} \left(x_{2n} - x_{2n-2} \right) + x_0 \end{align*} $$ So, applying Stolz-Cesàro (which I fortunately know about), we get that $\frac{x_{2N+1}}{N} \to 0$ and $\frac{x_{N}}{N} \to 0$. Both these statements can be bundled back together as: $\frac{x_n}{n} \to 0$, which is enough to imply the result. (The hint that I supplied is slightly prettier, but it's the same idea.)

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The result follows from the fact that $\frac{x_n}{n}\to0$, which itself follows from the facts that $\frac{x_{2n}}{n}\to0$ and $\frac{x_{2n+1}}{n}\to0$.

To show that $\frac{x_{2n}}{n}\to0$, write $x_{2n}=x_0+\sum\limits_{k=1}^ny_k$ with $y_k=x_{2k}-x_{2k-2}$. Then $y_n\to0$ by hypothesis hence the Cesàro means $C_n(y)=\frac1n\sum\limits_{k=1}^ny_k$ are such that $C_n(y)\to0$. Since $\frac{x_{2n}}{n}=\frac{x_{0}}{n}+C_n(y)$, this shows that $\frac{x_{2n}}{n}\to0$. The proof that $\frac{x_{2n+1}}{n}\to0$ is similar, using $x_{2n+1}=x_1+\sum\limits_{k=1}^nz_k$ with $z_k=x_{2k+1}-x_{2k-1}$.

Finally, to show that $t_n\to0$ implies $C_n(t)\to0$, the standard epsilon-delta method applies.

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