find the sum
1+1/√2 +1/√3+1√4------1/√n where n is perfect square
also find greatest integer value of (1+1/√2 +1/√3+1√4------1/√n ) where n is perfect square
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Let $\displaystyle S_n = \sum_{r=1}^n\frac1{\sqrt r}$. Then $\displaystyle S_n = \sqrt n\Bigl(\frac1n\sum_{r=1}^n\frac1{\sqrt {r/n}}\Bigr)$, and the expression in parentheses is a lower Riemann sum for $\displaystyle \displaystyle\int_0^1\!\!\!x^{-1/2}dx = 2$. Therefore $\displaystyle S_n<2\sqrt n$. On the other hand, the expression in parentheses is an upper Riemann sum for $\displaystyle \displaystyle\int_{1/n}^{(n+1)/n}\!\!\!x^{-1/2}dx$, which leads to the estimate $\displaystyle S_n>2\sqrt{n+1}-2$.
That's as close as I can get to pinning down the value of $\displaystyle S_n$. Its integer part (when n is a perfect square) must be either $\displaystyle 2\sqrt n-2$ or $\displaystyle 2\sqrt n-1$.
Second attempt at this problem. We are interested in the sum $\displaystyle S_n = \sum_{r=1}^n\frac1{\sqrt r}$ when $\displaystyle n$ is a perfect square, say $\displaystyle n = k^2$. Then the difference between successive sums of this form is
$\displaystyle \displaystyle S_{(k+1)^2} - S_{k^2} = \sum_{r=1}^{2k+1}\frac1{\sqrt{k^2+r}} = \sqrt{2k+1}\biggl(\frac1{2k+1}\sum_{r=1}^{2k+1}\fr ac1{\sqrt {\alpha +\frac r{2k+1}}}\biggr),$
where $\displaystyle \alpha = k^2/(2k+1)$. As in the previous comment, this is a lower Riemann sum for
$\displaystyle \displaystyle\sqrt{2k+1}\int_0^1\!\!\!(\alpha+x)^{-1/2}dx = \sqrt{2k+1}\Bigl[2(\alpha+x)^{1/2}\Bigr]_0^1 = 2$
(because $\displaystyle \sqrt{\alpha+1} - \sqrt\alpha = \frac{k+1}{\sqrt{2k+1}} - \frac k{\sqrt{2k+1}} = \frac1{\sqrt{2k+1}}$). Therefore $\displaystyle S_{(k+1)^2} - S_{k^2} < 2$.
The aim now is to show by induction that the integer part $\displaystyle \lfloor S_{k^2}\rfloor$ of $\displaystyle S_{k^2}$ is $\displaystyle 2k-2$. The base case is k=2, and for that you need to verify that $\displaystyle S_4\approx2.78$, so that $\displaystyle \lfloor S_{4}\rfloor = 2$.
For the inductive step, suppose that $\displaystyle \lfloor S_{k^2}\rfloor = 2k-2$. The above argument shows that $\displaystyle S_{(k+1)^2}<S_{k^2} + 2$, so that $\displaystyle \lfloor S_{(k+1)^2}\rfloor \leqslant \lfloor S_{k^2}\rfloor + 2 = 2(k+1)-2$. The reverse inequality $\displaystyle \lfloor S_{(k+1)^2}\rfloor \geqslant 2(k+1) - 2$ comes from the previous comment, where I showed that $\displaystyle S_n > 2\sqrt n - 2$.
Therefore the integer part of $\displaystyle S_n$, when n is a perfect square greater than 1, is given by $\displaystyle \lfloor S_n\rfloor = 2\sqrt n - 2$.