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,


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}

  • 1
    $\begingroup$ 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


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.

  • $\begingroup$ 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

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 $x_n<x_{n+1}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.