# Bounding sequence for $\left(1-\frac{n}{k}\right)^{\left\lfloor n/k\right\rfloor},\quad k,n\in\mathbb{N}.\quad k$ fixed

$$x_n=\left(1-\frac{k}{n}\right)^{\left\lfloor n/k\right\rfloor}$$

where $n\in\mathbb{N}$ and $k\in\mathbb{N}$ is fixed?

The problem is that the floor function at the exponent creates a rather erratic behaviour in the sequence (sawtooth-like). Ideally, a good bounding sequence would be quite tight, but looking at some graphs of this function I reckon that it may not be possible to achieve that.

So far I've only been able to derive some lower bounds, so any help with this would be greatly appreciated!

Thanks!

EDIT: I don't need a bounding sequence to deal with the limit for $n$ large, which is easy to study. In fact, I need a bounding sequence to make precise statements about all values of the sequence $x_n$, especially for $n$ small, that is when the behaviour is more erratic.

• $x_n$ bounds $x_n$, somehow, I don't think that's what you want. – user2345215 Mar 7 '14 at 19:11

Well, consider the function $f(x) = \left(1-\frac{k}{x}\right)^{\left\lfloor\frac{x}{k}\right\rfloor}$. What you're trying to find is $lim_{x\to \infty}f(x)$, aren't you? In this case $lim_{x\to \infty}f(x)=lim_{x\to \infty}\left(1-\frac{k}{x}\right)^{\left\lfloor\frac{x}{k}\right\rfloor}$ now let $\frac{k}{x} = \frac{1}{a}$ and you have$$lim_{x\to \infty}\left(1-\frac{k}{x}\right)^{\left\lfloor\frac{x}{k}\right\rfloor} = lim_{a\to \infty}\left(1-\frac{1}{a}\right)^{\left\lfloor\frac{1}{a}\right\rfloor} \leq lim_{a\to \infty}\left(1-\frac{1}{a}\right)^{\frac{1}{a}} =\frac{1}{e}$$ so the upper bounding to the sequence $$limsup[x_n]=limsup[\left(1-\frac{k}{n}\right)^{\left\lfloor\frac{n}{k}\right\rfloor}]=\frac{1}{e}$$
• I'm sorry, I should have made clearer in my question that I don't really need a bounding sequence to compute a limit, and in fact I'm actually interested in the "small $n$" behaviour, which is were non-monotonicity is more evident. I'll edit the question. – LErgodic Mar 8 '14 at 14:55