# Is the sequence $x_n=\dfrac{1}{\sqrt{n}}\left(1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\ldots+\dfrac{1}{\sqrt{n}}\right)$ monotone?

Observe that $$x_1=1$$ and $$x_2=\dfrac{1}{\sqrt{2}}\left(1+\dfrac{1}{\sqrt{2}}\right)>\dfrac{1}{\sqrt{2}}\left(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}\right)=1$$.

Thus, $$x_2>x_1$$. In general, we also have $$x_n=\dfrac{1}{\sqrt{n}}\left(1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\ldots+\dfrac{1}{\sqrt{n}}\right)>\dfrac{1}{\sqrt{n}}\left(\dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n}}+\ldots+\dfrac{1}{\sqrt{n}}\right)=1$$.

Thus, $$x_n\geq 1$$ for all $$n\in \mathbb{N}$$. Also, we have,

$$x_{n+1}=\dfrac{1}{\sqrt{n+1}}\left(\sqrt{n}x_n+\dfrac{1}{\sqrt{n+1}}\right)=\dfrac{\sqrt{n}}{\sqrt{n+1}}x_n+\dfrac{1}{n+1}$$.

Is it true that $$x_{n+1}>x_n$$?

Edit : Thanks to the solution provided by a co-user The73SuperBug. Proving $$x_{n+1}-x_n>0$$ is equivalent to proving $$x_n<1+\dfrac{\sqrt{n}}{\sqrt{n+1}}$$. This is explained below :

\begin{aligned} &x_{n+1}-x_n>0\\ \Leftrightarrow & \dfrac{\sqrt{n}}{\sqrt{n+1}}x_n+\dfrac{1}{n+1}-x_n>0\\ \Leftrightarrow & \left(1-\dfrac{\sqrt{n}}{\sqrt{n+1}}\right)x_n-\dfrac{1}{n+1}<0\\ \Leftrightarrow & x_n<\dfrac{1}{(n+1)\left(1-\dfrac{\sqrt{n}} {\sqrt{n+1}}\right)}\\ \Leftrightarrow & x_n<1+\dfrac{\sqrt{n}}{\sqrt{n+1}}. \end{aligned}

• It's basically a Riemann sum for a convex function. The (good) accepted answer focuses on the specific function you have, but I think a generalization would not harm, so I've added an answer based on this alternative idea. May 17, 2020 at 12:17

The sequence is indeed increasing. Using what you have left off we need to prove: $$x_{n+1} - x_n > 0\iff ...x_n < 1+\dfrac{\sqrt{n}}{\sqrt{n+1}}$$. We prove this by induction on $$n \ge 1$$. Clearly $$x_1 = 1 < 1+ \sqrt{\frac{1}{2}}$$. Assume $$x_n < 1+\dfrac{\sqrt{n}}{\sqrt{n+1}}$$, we show: $$x_{n+1} < 1+\dfrac{\sqrt{n+1}}{\sqrt{n+2}}$$. Using the recursive formula you had above: $$x_{n+1} = \dfrac{\sqrt{n}}{\sqrt{n+1}}x_n+\dfrac{1}{n+1}< \dfrac{\sqrt{n}}{\sqrt{n+1}}\left(1+\dfrac{\sqrt{n}}{\sqrt{n+1}}\right)+\dfrac{1}{n+1}= 1+\dfrac{\sqrt{n}}{\sqrt{n+1}}< 1+\dfrac{\sqrt{n+1}}{\sqrt{n+2}}$$ which is clear because $$n(n+2) < (n+1)^2$$. Thus by induction $$x_n < 1 +\dfrac{\sqrt{n}}{\sqrt{n+1}}$$ and in turn implies $$x_{n+1} > x_n, \forall n \ge 1$$. Thus the sequence is increasing.
A generalization I couldn't pass by. Let $$\color{blue}{S_n=\frac1n\sum_{k=1}^{n-1}f\big(\frac{k}{n}\big)}$$ where $$f:(0,1)\to\mathbb{R}$$ is strictly convex: $$f\big((1-t)a+tb\big)<(1-t)f(a)+tf(b)\quad\impliedby\quad a If we put $$a=k/(n+1),b=(k+1)/(n+1),t=k/n$$ for $$0 here, we obtain $$f\Big(\frac{k}{n}\Big)<\Big(1-\frac{k}{n}\Big)f\Big(\frac{k}{n+1}\Big)+\frac{k}{n}f\Big(\frac{k+1}{n+1}\Big),$$ which, after summing over $$k$$ (to have "$$<$$" still, we must assume $$n>1$$), gives $$\sum_{k=1}^{n-1}f\Big(\frac{k}{n}\Big)<\sum_{k=1}^{\color{red}{n}}\Big(1-\frac{k}{n}\Big)f\Big(\frac{k}{n+1}\Big)+\sum_{k=\color{red}{1}}^{n}\frac{k-1}{n}f\Big(\frac{k}{n+1}\Big)=\frac{n-1}{n}\sum_{k=1}^{n}f\Big(\frac{k}{n+1}\Big),$$ i.e. $$\color{blue}{n^2 S_n<(n^2-1)S_{n+1}}$$. Returning to the question, if we put $$f(x)=1/\sqrt{x}$$, we get $$x_n=S_n+1/n$$ and $$n^2 x_n<(n^2-1)x_{n+1}+1$$; since $$x_{n+1}>1$$, the latter implies the needed $$x_n.