Two interlaced by inequalities sequences: arithmetic and geometric
问题内容
We are to prove that if $n$ is fixed natural number then exists arithmetic $a_n$ and geometric $b_n$ sequences of integers that: $$b_1 < a_1 < b_2 < a_2 < \ldots < b_n < a_n.$$
Sketch
It’s equivalent to construct such sequences of fractions – we can always multiply by such large $N$ as is needed.
Let’s see an example: $n=3$ and $$1< 3 <4 < 11 < 16 < 19,$$ but it can’t be expanded for $n=4$: next terms are $64$, $27\ldots$ And we put $$\begin{cases} a_k = 1+ (k+1)c \\ b_k =(1+c)^{k+1} \end{cases},$$ where $c>0$ is small enough.
Then $a_{k-1} < b_k$ by means of Bernoulli inequality.
I’m keen on seeing another techniques of prove this theorem.
回答 (1)
Here's a construction: $$a_k = n^{2n} + (k+1)n^{2n-2}$$$$b_k = {n^{2n-2k}(n^2+1)^k}$$ Clearly $a_k$ is arithmetic and $b_k$ is geometric by calculating $b_{k+1}/b_k$
Then, $$\frac{b_k}{n^{2n}} = \left(1 + \frac{1}{n^2}\right)^k$$$$ \frac{a_k}{n^{2n}} = 1 + \frac{k+1}{n^2}$$ Bernoulli immediately gives $a_k < b_{k+1}$ and $b_k<a_k$ follows from:$$\begin{aligned} \left(1 + \frac{1}{n^2}\right)^k &= 1 + \frac{k}{n^2} + \sum_{j=2}^{k} \binom{k}{j} \frac{1}{n^{2j}} \\ &\le 1 + \frac{k}{n^2} + \frac{1}{n^2} \sum_{j=2}^{\infty} \frac{1}{j!} \\ &< 1 + \frac{k+1}{n^2} \end{aligned}$$