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I showed this inequality by induction. I want other methods to prove it. $(\frac{2n}{3}+\frac{1}{3})\sqrt{n}\leq \sum_{k=1}^{n}\sqrt{k}\leq (\frac{2n}{3}+\frac{1}{2})\sqrt{n}$

Thank

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4 Answers 4

up vote 0 down vote accepted

For $0\le k\le n-1$, let $R_k$ be the region defined by $k\le x\le k+1$, $\sqrt{x}\le y\le \sqrt{k+1}$,

so its area is given by

$A_k=\sqrt{k+1}\cdot1-\int_{k}^{k+1}\sqrt{x}\;dx=\sqrt{k+1}-\frac{2}{3}((k+1)^{3/2}-k^{3/2})$.

$\textbf{1)}$ Since the graph of $y=\sqrt{x}$ is concave down, $R_k$ is contained in the triangle with vertices $\;\;\;(k,\sqrt{k}), (k,\sqrt{k+1}), (k+1,\sqrt{k+1})$, so

$\;\;\;\sqrt{k+1}-\frac{2}{3}((k+1)^{3/2}-k^{3/2})\le\frac{1}{2}(\sqrt{k+1}-\sqrt{k})$.

Then $\displaystyle \sum_{k=0}^{n-1}[\sqrt{k+1}-\frac{2}{3}((k+1)^{3/2}-k^{3/2})]\le\sum_{k=0}^{n-1}\frac{1}{2}(\sqrt{k+1}-\sqrt{k})$, so

$\displaystyle\sum_{k=0}^{n-1}\sqrt{k+1}-\frac{2}{3}n^{3/2}\le\frac{1}{2}\sqrt{n}$ and therefore

$\displaystyle \sum_{k=1}^{n}\sqrt{k}\le\frac{2}{3}n^{3/2}+\frac{1}{2}\sqrt{n}=\big(\frac{2}{3}n+\frac{1}{2}\big)\sqrt{n}$.

$------------------------------------------$

$\textbf{2)}$ Since $(4k^2+4k+1)k\ge4k^3+4k^2$, $(2k+1)^2k\ge4k^2(k+1)$

$\;\;\;$so $(2k+1)\sqrt{k}\ge2k\sqrt{k+1}\;\;\;$. Then

$\;\;\;2k\sqrt{k}+\sqrt{k}+3\sqrt{k+1}\ge2(k+1)\sqrt{k+1}+\sqrt{k+1}$, so

$\;\;\;\frac{2}{3}k\sqrt{k}+\frac{1}{3}\sqrt{k}+\sqrt{k+1}\ge\frac{2}{3}(k+1)\sqrt{k+1}+\frac{1}{3}\sqrt{k+1}$ and

$\;\;\;\sqrt{k+1}-\frac{2}{3}(k+1)\sqrt{k+1}+\frac{2}{3}k\sqrt{k}\ge\frac{1}{3}\sqrt{k+1}-\frac{1}{3}\sqrt{k}$.

Then $\sqrt{k+1}-\frac{2}{3}((k+1)^{3/2}-k^{3/2})\ge\frac{1}{3}(\sqrt{k+1}-\sqrt{k})$, so

$\displaystyle\sum_{k=0}^{n-1}[\sqrt{k+1}-\frac{2}{3}((k+1)^{3/2}-k^{3/2})]\ge\sum_{k=0}^{n-1}\frac{1}{3}(\sqrt{k+1}-\sqrt{k})$.

Then $\displaystyle\sum_{k=0}^{n-1}\sqrt{k+1}-\frac{2}{3}n^{3/2}\ge\frac{1}{3}\sqrt{n}$, and therefore

$\displaystyle\sum_{k=1}^{n}\sqrt{k}\ge\frac{2}{3}n^{3/2}+\frac{1}{3}\sqrt{n}=\big(\frac{2}{3}n+\frac{1}{3}\big)\sqrt{n}$.

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Hint

By these inequalities (using the fact that the function $x\mapsto \sqrt x$ is increasing) $$\sqrt k\le\int_k^{k+1}\sqrt x dx=\frac 2 3\left((k+1)^{3/2}- k^{3/2}\right)\le\sqrt{k+1}$$ and we sum by telescoping we get the result.

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I get $\frac{2}{3}n\sqrt{n}\leq \sum_{k=1}^{n}\sqrt{k}\leq \frac{2}{3}((n+1)\sqrt{n+1}-1)$ –  yoda Jan 5 at 19:54

($\frac{2n}{3}$+$\frac{1}{3})$$\sqrt{n}$$\leq $$\sum_{k=1}^{n}$$\sqrt{k}$$\leq $$(\frac{2n}{3}$+$\frac{1}{2})$$\sqrt{n}$ ?

By these inequalities :
$\forall$ k $\geq$ 1 : $\frac{2}{3}$[$k^\frac{3}{2}$-$(k-1)^\frac{3}{2}$]+$\frac{1}{3}$[$k^\frac{1}{2}$-$(k-1)^\frac{1}{2}$]$\leq$ $k^{\frac{1}{2}}$$\leq$ $\frac{2}{3}$[$k^{\frac{3}{2}}$-$(k-1)^{\frac{3}{2}}$]+$\frac{1}{2}$[$k^{\frac{1}{2}}$-$(k-1)^{\frac{1}{2}}$] $(*)$
and we sum by telescoping we get the result.

show (*)

3$a_{k}$=2[$k^{\frac{3}{2}}$-$(k-1)^{\frac{3}{2}}$]+[$k^{\frac{1}{2}}$-$(k-1)^{\frac{1}{2}}$] - 3$k^{\frac{1}{2}}$= 2$k^{\frac{1}{2}}$(k-1)-$(k-1)^{\frac{1}{2}}(2(k-1)+1)$ $\\$ 3$a_{k}$=2$k^{\frac{1}{2}}$(k-1)-$(k-1)^{\frac{1}{2}}$(2k-1)$\leq $0 because 4k$(k-1)^{2}$$\leq$ (k-1)$(2k-1)^{2}$ indeed 4k(k-1)-(2k-1)²= -1$\leq $0

6$b_{k}$=6$k^{\frac{1}{2}}$-4[$k^{\frac{3}{2}}$-$(k-1)^{\frac{3}{2}}$]-3[$k^{\frac{1}{2}}$-$(k-1)^{\frac{1}{2}}$]
6$b_{k}$=$(k-1)^{\frac{1}{2}}$(4k-1)-$k^{\frac{1}{2}}$(-3+4k)$\leq$ 0 because (k-1)$(4k-1)^{2}$$\leq$ k$(4k-3)^{2}$
indeed (k-1)$(4k-1)^{2}$ - k$(4k-3)^{2}$= -1$\leq$0

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For the upper bound use $$ \int_{k-1}^{k} \sqrt x \, dx \ge \frac12(\sqrt{k-1} + \sqrt{k}) .$$ You can see this by drawing a picture, and looking at the area under the trapezoid.

I don't have the lower bound yet. But I am hoping that a more careful estimate of the area might do the trick.

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