Limit of product related to a cv sequence If $(a_n)$ is a convergent sequence, can one assert that the product $\displaystyle\prod_{k=1}^n(a_{n+k}-a_k) $ converges to $0$ ?
If so, how can one prove this ?
Any ideas are welcome, thanks.
 A: $ \frac 1 n \sum_{k=1}^{n} \log |a_{n+k}-a_k| \to -\infty $ (see details below) so $\sum  \log |a_{n+k}-a_k| \to -\infty $ ; taking exponential $\prod_{k=1}^{n} |a_{n+k}-a_k| \to 0$. To see why $ \frac 1 n \sum_{k=1}^{n} \log |a_{n+k}-a_k| \to -\infty $ note that $\log t \to -\infty $ as $t$ decreases to $0$. Given any integer $N$ there exists $m$ such that $\log |a_{n+k}-a_k| <-N$ for $k \geq m$. If you split $\sum_{k=1}^{n} \log |a_{n+k}-a_k|$ into sum from $1$ to $m$ and $m+1$ to $n$ it should be clear that $ \frac 1 n \sum_{k=1}^{n} \log |a_{n+k}-a_k| \to -\infty $ since the first part goes to $0$ and the second part to $-\infty $.
A: Sketch:


*

*Since the sequence $(a_n)$ is convergent, it is bounded.  Therefore, there exists an $M>0$ such that for all $i$ and $j$, $|a_i-a_j|<M$.

*Since the sequence $(a_n)$ is convergent, it is Cauchy.  Therefore, there exists an $N>0$ such that if $i$ and $j$ are greater than $N$, $|a_i-a_j|<\frac{1}{2}$.

*Putting this together, for $n$ sufficiently large,
$$
\prod_{k=1}^n(a_{n+k}-a_k)=\prod_{k=1}^N(a_{n+k}-a_k)\prod_{k=N+1}^n(a_{n+k}-a_k).
$$ 
Taking absolute values, we get
$$
\left|\prod_{k=1}^n(a_{n+k}-a_k)\right|=\prod_{k=1}^n\left|a_{n+k}-a_k\right|=\prod_{k=1}^N\left|a_{n+k}-a_k\right|\prod_{k=N+1}^n\left|a_{n+k}-a_k\right|\leq \frac{M^N}{2^{n-N}}.
$$


By allowing little $n$ to grow, you see that $M$ and $N$ stay constant, so the absolute value of the product of interest approaches $0$, so the product itself approaches $0$.
