<s>[INST] Limit involving power tower: $\lim\limits_{n\to\infty} \frac{n+1}n^{\frac n{n-1}^\cdots}$ What is the value of the following limit?
$$\large \lim_{n \to \infty} \left(\frac{n+1}{n}\right)^{\frac{n}{n-1}^{\frac{n-1}{n-2}^{...}}}$$ In general what do limits of infinite decreasing numbers strung together in familiar ways approach?
 [/INST] For $n\ge 1$, let $$a_n=\left(\frac{n+1}{n}\right)^{\frac{n}{n-1}^{\frac{n-1}{n-2}^{...}}}$$ where the tower stops when we reach $2/1=2$, so $a_1=2$ and $$a_{n+1}=\left(\frac{n+2}{n+1}\right)^{a_n}.$$
Note that each $a_n$ is (strictly) larger than $1$, and that the sequence is decreasing from $n=2$ on: First, $a_2=9/4>2>1.911>a_3$. Next, if $a_{n+1}<a_n$, then $$a_{n+2}=\left(\frac{n+3}{n+2}\right)^{a_{n+1}}<\left(\frac{n+2}{n+1}\right)^{a_{n+1}}<\left(\frac{n+2}{n+1}\right)^{a_n}=a_{n+1}. $$
Since the sequence is decreasing and bounded below by $1$, it follows that $\lim_n a_n=a$ exists, and satisfies $a\ge 1$.
It remains to argue that $a=1$. To see this, note that $a_{n+1}=b_n^{a_n}$, where $\displaystyle b_n=\frac{n+2}{n+1}$, so $a=\lim_n a_{n+1}=(\lim_n b_n)^{\lim_n a_n}=1^a=1$.
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