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How do i evaluate this limit : $$\displaystyle \lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+............+\frac{1}{\sqrt{n}}\right)$$ ?

Thank you for any help .

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marked as duplicate by quid, Namaste calculus Sep 25 '17 at 0:43

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ may you mean \frac {1}{(sqrt(n)} in LHS in the title ? $\endgroup$ – zeraoulia rafik Jul 4 '15 at 14:46
  • $\begingroup$ It is bounded above by 2 so it certainly has a limit. $\endgroup$ – Paul Jul 4 '15 at 14:55
  • $\begingroup$ also should you have $\frac{1}{\sqrt{k}}$ $\endgroup$ – Chinny84 Jul 4 '15 at 16:16
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Let $a_n = 1 + \dfrac {1}{\sqrt{2}} + \cdots + \dfrac {1}{\sqrt{n}}$ and $b_n = \sqrt{n}$ for all $n\geq 1$. We have $$\lim_{n\to \infty} \dfrac {a_{n+1} - a_n} {b_{n+1} - b_n} = \lim_{n\to \infty} \dfrac {1} {\sqrt{n+1}(\sqrt{n+1} - \sqrt{n})} = \lim_{n\to \infty} \dfrac {\sqrt{n+1} + \sqrt{n}} {\sqrt{n+1}} = 2,$$ so by the Stolz-Cesàro theorem we get $\lim_{n\to \infty} \dfrac {a_n} {b_n} = 2$

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  • $\begingroup$ it is a nice answer $\endgroup$ – sproot.jack novel Jul 4 '15 at 14:59
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\begin{align} &\frac{1}{\sqrt{n}}\left(\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \ldots + \frac{1}{\sqrt{n}}\right) \\ = \ &\frac{1}{n}\left(\sqrt{\frac{n}{1}} + \sqrt{\frac{n}{2}} + \ldots + \sqrt{\frac{n}{n}}\right) \\ = \ &\frac{1}{n}\left(\frac{1}{\sqrt{\frac{1}{n}}} + \frac{1}{\sqrt{\frac{2}{n}}} + \ldots + \frac{1}{\sqrt{\frac{n}{n}}}\right) \\ \rightarrow \ &\int_0^1 \frac{1}{\sqrt{x}} \mathrm{d}x = 2 \end{align}

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We bound $s_n=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}}$ by integrals. We get $$\int_1^n \frac{dt}{\sqrt{t}}\le s_n\le 1+\int_1^{n-1} \frac{dt}{\sqrt{t}}.$$ Ealuate the integrals, divide by $\sqrt{n}$, and Squeeze.

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