Monotone sequence

Start with $\;\displaystyle \frac{n}{2a_{n+1}} = a_1+\ldots+a_{n-1}+a_n = (a_1+\ldots+a_{n-1})+a_n = \frac{n-1}{2a_{n}}+a_n\,$, then:

• $\;\displaystyle \frac{1}{a_{n+1}} = \frac{n-1}{na_{n}}+\frac{2a_n}{n} \ge \sqrt{2}\,$ by induction for $\,n \ge 2\,$
  because $\,2a_n^2-\sqrt{2}n a_n + n-1$ $\displaystyle=2 \left(a_n - \frac{1}{\sqrt{2}}\right)\left(a_n - \frac{n-1}{\sqrt{2}}\right)$ $\ge 0\,$ when $\displaystyle\,a_n \le \frac{1}{\sqrt{2}}\,$

• $\;\displaystyle \frac{1}{a_{n+1}} - \frac{1}{a_{n}} = -\frac{1}{na_{n}}+\frac{2a_n}{n} \le 0\,$ since $\displaystyle\,a_n^2 \le \frac{1}{2}\,$, and therefore$\,a_{n+1} \ge a_n\,$