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Let $a_n=\frac{(n+1)^{n+2}}{n^n}$ and $b_n=\frac{(n+2)^{(n+1)}}{(n+1)^{n-1}}$. Then it is easy to see that $a_n \leq b_n$ for all integers $n\geq 1$ (because the sequence $(1+\frac{1}{n})^n$ is increasing).

My question is : is there always an integer between $a_n$ and $b_n$ ? This holds for $1 \leq n \leq 100$.

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    $\begingroup$ Since the limit of $b(n)-a(n)$ is $\frac{1}{2}e > 1$, this must holds for $n$ large too. $\endgroup$ – Bernikov Dec 12 '11 at 10:08
  • $\begingroup$ Experimentally $b_n - a_n$ is increasing. $\endgroup$ – joro Dec 12 '11 at 10:37
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Yes, the difference between $b_n$ and $a_n$ is always at least $1$. Let $$f(n)=(1+1/n)^n,$$ so that $a_n=(n+1)^2f(n)$ and $b_n=(n+1)^2f(n+1).$ Then by the mean value theorem we have that $$b_n-a_n=(n+1)^2(f(n+1)-f(n))=(n+1)^2f'(c)$$ for some $c\in (n,n+1)$. Next we calculate $$f'(x)=f(x)\left(\log(1+1/x)-\frac{1}{x(1+1/x)}\right),$$ and after substituting in the power series expansions for the functions in the parenthesis we have $$f'(x)=f(x)\left(\frac{1}{2x^2}-\frac{2}{3x^3}+\frac{3}{4x^4}-\cdots\right).$$ If $x>2$ then the terms in this series are decreasing in absolute value and $$f'(x)\ge \frac{f(x)}{2x^2}\left(1-\frac{4}{3x}\right).$$ Keeping in mind that $f(x)$ is increasing, it is easy to check that this expression is greater than $1/x^2$ whenever $x>7$. Substituting this back in the above equation gives $b_n-a_n>1$ for all $n>7$, and the rest is verified directly.

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