Strictly decreasing sequence Let $$a_n=\sum_{k=1}^n\sqrt{k^2+1}$$
Define the sequence $b_n=\dfrac{2a_n}{n} - n$.
I need to show that the sequence $\langle b_n\rangle_{n=1}^\infty$ is strictly decreasing.

Try to show $b_{n+1} - b_n < 0$:
$b_{n+1} - b_n = \left[\dfrac{2a_{n+1}}{n+1} - (n+1)\right] -  \left[\dfrac{2a_n}n - n\right]$
$ = \dfrac1{n(n+1)}\left[2n \sum_{k=1}^{n+1}\sqrt{k^2+1} - 2(n+1) \sum_{k=1}^n\sqrt{k^2+1} - n(n+1)\right]$
$ = \dfrac1{n(n+1)} \left[2n \sqrt{(n+1)^2 + 1} - 2 \sum_{k=1}^n\sqrt{k^2+1} - n(n+1)\right]\qquad\qquad\mbox{(1)}$
$ < \dfrac1{n(n+1)} \left[2n \sqrt{\left(n + 1 + \dfrac1{2n}\right)^2} - 2 \sum_{k=1}^n \left(k + \dfrac1{2k}\right) - n(n+1)\right]$
$ < \dfrac1{n(n+1)} \left[2n \left(n + 1 + \dfrac1{2n}\right) - n(n+1) - \ln(n+1) - n(n+1)\right]$ ... [Ref: http://en.wikipedia.org/wiki/Harmonic_series_(mathematics)#Integral_test to show $\sum_{k=1}^n\dfrac1k > \ln(n+1)$]
$ = \dfrac1{n(n+1)} \left[1 - \ln(n+1)\right]$
$ < 0$ for all $n \geq 2$.
For $n = 1$ case: calculate: $b_2 - b_1 = \sqrt5 - \sqrt2 - 1 < 0$.
Revised part to fix mistake GL5 found, following from (1):
$ < \dfrac1{n(n+1)} \left[2n \sqrt{\left(n + 1 + \dfrac1{2(n+1)}\right)^2} - 2 \sum_{k=1}^n \left(k + \dfrac1{2(k+1)}\right) - n(n+1)\right]$
$ < \dfrac1{n(n+1)} \left[2n \sqrt{\left(n + 1 + \dfrac1{2(n+1)}\right)^2} - n(n+1) -  \sum_{k=2}^{n+1} \dfrac1{k} - n(n+1)\right]$
$ < \dfrac1{n(n+1)} \left[2n \left(n + 1 + \dfrac1{2(n+1)}\right) - n(n+1) - \ln(n+1) + 1 - \dfrac1{n+1} - n(n+1)\right]$
$ = \dfrac1{n(n+1)} \left[2 - \dfrac2{n+1} - \ln(n+1)\right]$
$ < 0$ for all $n \geq 4$.
Initial cases for $1 \leq n \leq 3$ to be done by individual calculation. Having 3 initial cases is inelegant. There might be a way to tighten up the inequalities to reduce it.
Mick
