Limit involving power tower: $\lim\limits_{n\to\infty} \frac{n+1}n^{\frac n{n-1}^\cdots}$

Solution 1:

The sequence is given by:

$$a_1 = 2, a_n = \left(\frac{n+1}n\right)^{a_{n-1}}$$

Taking logs, we obtain:

$$\log a_n = a_{n-1} \log\left(1+\frac1n\right)$$

and it is easy to show that $\dfrac1{2n} \le \log\left(1+\frac1n\right) \le \dfrac1n$.

Now if we can show that $a_n$ is bounded, we are done by the Squeeze theorem (since then $\lim\limits_{n\to\infty} \log a_n = 0$, hence $\lim\limits_{n\to\infty} a_n = 1$).

Obviously, $a_n \ge 0$ for all $n$. We prove inductively that $a_n \le e$. The basis is trivial: $a_1 = 2 \le e$. Suppose $a_{n-1} \le e$. By the above estimate, $\log a_n \le \dfrac en \le 1$ for $n \ge 3$. It remains to show that $a_2 \le e$:

$$a_2 = \left(\dfrac32\right)^2 = \dfrac94 \le e$$

In conclusion:

$$\lim_{n\to\infty} a_n = \lim_{n \to \infty} {\large\frac{n+1}{n}^{\frac{n}{n-1}^{\frac{n-1}{n-2}^{...}}}} = 1$$

Solution 2:

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 $a=\lim_n a_{n+1}=(\lim_n b_n)^{\lim_n a_n}=1^a=1$.