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Calculate $\lim\limits_{n\to\infty} V_{n}$ ($\forall n\in \mathbb{N}$), $V_{n}=\left(\sum\limits_{k=0}^{2n}\sqrt{k}\right)-(2n+1)\sqrt{n}$.

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What have you tried yourself? –  Link Dec 19 '13 at 22:06
    
I have tried to get sqrt(n) into the sum and then i multiplied with sqrt(k)+sqrt(n) then i tried to frame v_{n} but i couldn't find the answer –  user115947 Dec 19 '13 at 22:17

2 Answers 2

Dividing by $n$ seems to be just right. Let us note

$$a_n=\frac{V_n}{(2n+1)\sqrt n}=\frac{1}{(2n+1)}\sum_{k=0}^{2n}\sqrt{\frac kn}-1.$$

We see that $(a_n+1)_n$ is a Riemann sum and that it should converge toward the integral $$\frac12\int_0^2\sqrt{x}\,\mathrm{d}x=\frac{2\sqrt{2}}3.$$ Now we have $$V_n \underset{n\to\infty}{\sim} \left(\frac{2\sqrt2}3-1\right)(2n+1)\sqrt n.$$ As $\frac{2\sqrt2}3-1\simeq-0.057$ the limit is $-\infty$.

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The limit of $V'_n=\sum_{k=0}^{2n}\sqrt k - \frac23(2n+1)\sqrt{2n}$ seems much more interesting then... –  V. Rossetto Dec 19 '13 at 22:38

Hint: For every $k\geqslant1$, $$ \int_{k-1}^{k}\sqrt{x}\,\mathrm dx\leqslant\sqrt{k}\leqslant\int_k^{k+1}\sqrt{x}\,\mathrm dx.$$

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we have not reached this part in our lessons yet –  user115947 Dec 19 '13 at 22:24
    
I doubt that very much. Can you show that $\sqrt{x}\leqslant\sqrt{k}\leqslant\sqrt{z}$ for every $x$ in $(k-1,k)$ and every $z$ in $(k,k+1)$? –  Did Dec 19 '13 at 22:28
    
No i'm telling you the truth.And no i'm sorry i can't –  user115947 Dec 19 '13 at 22:37
    
You cannot show that $x\leqslant k\implies\sqrt{x}\leqslant\sqrt{k}$? Come... –  Did Dec 20 '13 at 7:56

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