<s>[INST] Inequalities showing monotonicity imply another inequality I came across a proof that shows that $\lim\limits_{n \rightarrow \infty} (1+ \frac{1}{n})^{n\cdot (\frac{1}{n+1} +\dots + \frac{1}{2n})} =2$
First it is shown that $((1+\frac{1}{n})^{n+1})_{n=1}^{\infty}$ is monotonically decreasing and $((1+\frac{1}{n})^n)_{n=1}^{\infty}$ is monotonically increasing, thus
$$(1 + \frac{1}{n})^n \leq (1 + \frac{1}{n+1})^{n+1} \leq \dots \leq (1+ \frac{1}{n+k})^{n+k}$$ for $((1+\frac{1}{n})^n)_{n=1}^{\infty}$ and
$$(1+\frac{1}{n})^{n+1} \geq (1+ \frac{1}{n+1})^{n+2} \geq \dots \geq (1 + \frac{1}{n+k})^{n+k+1}$$ for $((1+\frac{1}{n})^{n+1})_{n=1}^{\infty}$
I understand the proof up until this point. But then it says that those two inequalities yield
$$
\left(1+\frac1{n+k}\right)^{\frac{n}{n+1}}\le\left(1+\frac1n\right)^\frac{n}{n+k}\le\left(1+\frac1{n+k}\right)
$$
How do I obtain this result, is there something obvious I am missing?
 [/INST] By starting from
$$
   \left(1 + \frac1n\right)^{n} \le \left(1 + \frac1{n+k}\right)^{n+k}
$$
and raising both sides to the power $\frac1{n+k}$, we get
$$
   \left(1 + \frac1n\right)^{\frac{n}{n+k}} \le \left(1 + \frac1{n+k}\right). \tag{1}
$$
Also, by starting from
$$
   \left(1 + \frac1{n+k}\right)^{n+k+1} \le \left(1 + \frac1n\right)^{n+1} 
$$
and raising both sides to the power $\frac{n}{(n+1)(n+k+1)}$, we get
$$
   \left(1 + \frac1{n+k}\right)^{\frac{n}{n+1}} \le \left(1 + \frac1n\right)^{\frac{n}{n+k+1}}. \tag{2}
$$
Finally, we have
$$
   \left(1 + \frac1n\right)^{\frac{n}{n+k+1}} \le \left(1 + \frac1n\right)^{\frac{n}{n+k}} \tag{3}
$$
just because the exponent is bigger on the right.
To get the inequality we want, chain together $(2)$, $(3)$, and $(1)$.
</s>