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The question arise in connection with this problem

Prove that $$\lim_{n\rightarrow \infty}\sqrt{\frac{1}n}-\sqrt{\frac{2}n}+\sqrt{\frac{3}n}-\cdots+\sqrt{\frac{4n-3}n}-\sqrt{\frac{4n-2}n}+\sqrt{\frac{4n-1}n}=1$$

Thanks to answer @Vincenzo Oliva. I forgot the Stolz-Cesàro theorem

Roman83
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2 Answers2

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Note that the sequence you're interested in can be written as $a_n=\dfrac1{\sqrt{n}}\left(\sqrt{4n-1}+\sum\limits_{k=1}^{2n-1}\sqrt{2k-1}-\sqrt{2k}\right).$ Applying the Stolz-Cesàro theorem, we have \begin{align}\lim_{n\to\infty} a_n&=\lim_{n\to\infty}\frac{\sqrt{4n+3}-\sqrt{4n-1}+\sqrt{4n-1}-\sqrt{4n}+\sqrt{4n+1}-\sqrt{4n+2}}{\sqrt{n+1}-\sqrt{n}} \\ &= \lim_{n\to\infty} 2\frac{\sqrt{1+3/(4n)}-1+\sqrt{1+1/(4n)}-\sqrt{1+1/(2n)}}{\sqrt{1+1/n}-1} \\ &=\lim_{n\to\infty} \frac2n\frac{3/8+1/8-1/4}{1/(2n)}=1.\end{align}

Vincenzo Oliva
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We have \begin{align} &\sqrt{\frac{1}{n}} - \sqrt{\frac{2}{n}} + \sqrt{\frac{3}{n}} - \cdots + \sqrt{\frac{4n-3}{n}} - \sqrt{\frac{4n-2}{n}} + \sqrt{\frac{4n-1}{n}} \\ =& \sqrt{\frac{1}{n}} +\sum_{k=1}^{2n-1}(\sqrt{\frac{2k+1}{n}}-\sqrt{\frac{2k}{n}}) \\ =& \sqrt{\frac{1}{n}} + \frac{1}{\sqrt{n}}\sum_{k=1}^{2n-1}\frac{1}{\sqrt{2k+1}+\sqrt{2k}} \tag{1} \end{align} Moreover, \begin{align} \sum_{k=1}^{2n-1}\frac{1}{\sqrt{2k+1}+\sqrt{2k}} \leq \sum_{k=1}^{2n-1}\frac{1}{2\sqrt{2k}} = \frac{1}{2\sqrt{2}}\sum_{k=1}^{2n-1}\frac{1}{\sqrt{k}} < \frac{1}{2\sqrt{2}}\int_0^{2n}x^{-1/2}dx = \sqrt{n} \tag{2} \end{align} and \begin{align} \sum_{k=1}^{2n-1}\frac{1}{\sqrt{2k+1}+\sqrt{2k}} \geq \sum_{k=1}^{2n-1}\frac{1}{2\sqrt{2k+2}} = \frac{1}{2\sqrt{2}}\sum_{k=2}^{2n}\frac{1}{\sqrt{k}}>\frac{1}{2\sqrt{2}}\int_2^{2n}x^{-1/2}dx = \sqrt{n}-1 \tag{3} \end{align} Provided (2) and (3), we conclude that $$ \lim_{n\rightarrow \infty} \sqrt{\frac{1}{n}} + \frac{1}{\sqrt{n}}\sum_{k=1}^{2n-1}\frac{1}{\sqrt{2k+1}+\sqrt{2k}} = 1 $$

PSPACEhard
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  • why $$\sum_{k=2}^{2n}\frac{1}{\sqrt{k}}>\int_2^{2n}x^{-1/2}dx$$ and $$\sum_{k=1}^{2n-1}\frac{1}{\sqrt{k}}<\int_0^{2n}x^{-1/2}dx?$$ – Leox Aug 14 '16 at 07:57
  • @Leox Observe the graph of $x^{-1/2}$. $\frac{1}{\sqrt{k}}$ is the area of a rectange with height $\frac{1}{\sqrt{k}}$ and width $1$, while $\int_k^{k+1}x^{-1/2}dx$ is the area of the region below $x^{-1/2}$ on the interval $[k, k+1]$. Therefore, $\frac{1}{\sqrt{k}} > \int_k^{k+1} x^{-1/2}dx$ and thus $$\sum_{k=2}^{2n}\frac{1}{\sqrt{k}} > \sum_{k=2}^{2n} \int_{k}^{k+1}x^{-1/2}dx = \int_{2}^{2n+1}x^{-1/2}dx > \int_{2}^{2n} x^{-1/2}dx$$ The second inequality in your comment can be derived similarly. – PSPACEhard Aug 15 '16 at 05:36
  • Thank you. Mu question was - why we have different symbolls ( $>$ and $<$ ) for almost the same sums? – Leox Aug 15 '16 at 17:27