3
$\begingroup$

Show that $$\dfrac{\sqrt{2}}{3}n^2<\sqrt{1-\dfrac{\sqrt{1}}{n}}+\sqrt{1-\dfrac{\sqrt{2}}{n}}+\sqrt{1-\dfrac{\sqrt{3}}{n}}+\cdots+\sqrt{1-\dfrac{\sqrt{n^2-1}}{n}}<\sqrt{2}n^2.$$

Maybe use $$(1+x)^{\frac{1}{2}}<1+\dfrac{x}{2},x>-1$$ so $$\sqrt{1-\dfrac{\sqrt{k}}{n}}<1-\dfrac{1}{2}\cdot\dfrac{\sqrt{k}}{n}$$ then how to prove it?

I don't want see without Mathematical induction,maybe can use integral to prove it?

$\endgroup$
4
  • 1
    $\begingroup$ Mathematical induction may help you. $\endgroup$ Commented Aug 21, 2014 at 5:01
  • $\begingroup$ If I'm not wrong for your right inequality you have (by adding last inequality for all $k$) $n^2-1-(positive~ number)\lt n^2$. $\endgroup$
    – pointer
    Commented Aug 21, 2014 at 5:27
  • $\begingroup$ The title does not match the body; is it $\sqrt{n-1}$ or $\sqrt{n^2-1}$? $\endgroup$
    – vadim123
    Commented Aug 21, 2014 at 5:34
  • $\begingroup$ The form isn't perfect for it, but you're right - this looks suspiciously like a Riemann sum for an integral and I would expect that you can massage things to a point where you can treat it as such. $\endgroup$ Commented Aug 21, 2014 at 5:44

1 Answer 1

5
$\begingroup$

Note that $f(x)=\sqrt{1-\sqrt{x}}~$ is strictly decreasing on $[0,1]$. Thus for $0\leq k<m$ we have $$ \int_{k/m}^{(k+1)/m}f(t)dt<\frac{1}{m}f\left(\frac{k}{m}\right)\tag{1} $$ and for for $0< k\leq m$ we have $$ \int_{(k-1)/m}^{k/m}f(t)dt>\frac{1}{m}f\left(\frac{k}{m}\right)\tag{2} $$ So, if $$I_m=\frac{1}{m}\sum_{k=1}^{m-1}\sqrt{1-\sqrt{\frac{k}{m}}}$$ then, from $(1)$ we get $$ \int_{1/m}^1f(t)dt< I_m<\int_{0}^{1-1/m}f(t)dt<\int_{0}^{1}f(t)dt $$ Now, it is not difficult to see that $$\left(-\frac{4}{15}(1-\sqrt{x})^{3/2}(2+3\sqrt{x})\right)'=f(x)$$ Thus $$ \frac{4}{15}\left(1-\frac{1}{\sqrt{m}}\right)^{3/2}\left(2+ \frac{3}{\sqrt{m}}\right)<I_m<\frac{8}{15} $$ In particular, for $m=n^2$ we have $$ \frac{4}{15}\left(1-\frac{1}{n}\right)^{3/2}\left(2+ \frac{3}{n}\right)<I_{n^2}<\frac{8}{15} $$ Now, the left hand side of the previous inequality is an increasing function of $n$. So, for $n\geq 4$ we have $$ \frac{\sqrt{2}}{3}<\frac{11\sqrt{3}}{40}<I_{n^2}<\frac{8}{15} $$ Now, a direct check shows that the proposed inequality is valid also for $n=3$, but it is not for $n=1,2$. So, we have proved that So, for $n\geq 3$ we have $$ n^2\frac{\sqrt{2}}{3} <\sum_{k=1}^{n^2-1}\sqrt{1- \frac{\sqrt{k}}{n}}<\frac{8}{15}n^2 $$ which is the desired conclusion, with a stronger upper bound.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .