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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{equation} \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} \end{equation}

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    $\begingroup$ 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. $\endgroup$
    – metamorphy
    May 17, 2020 at 12:17

2 Answers 2

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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.

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  • $\begingroup$ Could you please explain the two way implication ? I am not clear with it $\endgroup$
    – Debashish
    May 17, 2020 at 4:27
  • $\begingroup$ Oh I got it ,.. thanks $\endgroup$
    – Debashish
    May 17, 2020 at 4:32
  • $\begingroup$ @Debashish: Glad to know you got it. I basically used what you had and continue... $\endgroup$
    – user781811
    May 17, 2020 at 4:33
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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: $$f\big((1-t)a+tb\big)<(1-t)f(a)+tf(b)\quad\impliedby\quad a<b,0<t<1.$$ If we put $a=k/(n+1),b=(k+1)/(n+1),t=k/n$ for $0<k<n$ 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 $x_n<x_{n+1}$.

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