I Got the set
$\sqrt[k]{k+1}$
I have found the supermum by the Inequality of arithmetic and geometric means. And the result is 2. I dont have a way for solving the infimum . I tried to solve it by move to to some eqaution and try by Binomial theorem, but i didnt success.
I know that the infimum is 1, but how can i proove it ?
Thanks.
On
To determine the supremum and infimum of the sequence of interest, we will first show that the sequence $a_n=(1+n)^{1/n}$ is monotonically decreasing for $n\ge 1$. We note that $a_n\ge a_{n+1}$ if and only if
$$(1+n)^{n+1}\ge (2+n)^n \tag 1$$
Now, forming the ratio $\frac{(1+n)^{n+1}}{(2+n)^n}$ reveals that
$$\begin{align} \frac{(1+n)^{n+1}}{(2+n)^n}&=\left(\frac{1+n}{2+n}\right)^{n+1}(2+n)\\\\ &=\left(1-\frac{1}{2+n}\right)^{n+1}(2+n) \tag 1\\\\ &\ge \left(1-\frac{n+1}{2+n}\right)(2+n) \tag 2\\\\ &=1 \end{align}$$
where in going from $(1)$ to $(2)$ we used Bernoulli's Inequality.
The preceding development shows that $(1+n)^{n+1}\ge (2+n)^n$, which is equivalent to $a_n\ge a_{n+1}$; thus $a_n$ is a monotonically decreasing sequence.
Since $a_n$ is decreasing, its maximum value is at $n=1$ for which $a_1=2$; therefore $a_1$ is the supremum of $a_n$.
Next, we find that
$$\lim_{n\to \infty} a_n =\lim_{n\to \infty} e^{\frac1n \log (1+n)}=1 \tag 3$$
Equation $(3)$ implies that for every $\epsilon>0$, there exists a number $N(\epsilon)$ such that whenever $n>N(\epsilon)$, $1-\epsilon<(1+n)^{1/n}<1+\epsilon$. Therefore, ,we have
$$\inf_{n\ge 1}(1+n)^{1/n}=1$$
and we are done!
Note that for any positive integer $k$, we have $\sqrt[k]{k+1}\gt 1$. We want to show that given any $\epsilon\gt 0$ we can find an $N$ such that $\sqrt[N]{N+1}\lt 1+\epsilon$.
By the Binomial Theorem, if $n\ge 2$ then $(1+\epsilon)^n \gt \frac{n^2}{2}\epsilon^2$. If $n$ is large enough, we therefore have $$(1+\epsilon)^n \gt n+1.$$ (What "large enough" means can be easily made explicit.)
Taking $n$-th roots, we conclude that for such an $n$ we have $$\sqrt[n]{n+1}\lt 1+\epsilon.$$