# $\mathop {\lim }\limits_{n \to \infty } {1 \over {\sqrt n }} \left({1 \over {\sqrt 1 }} + {1 \over {\sqrt 2 }} +\cdots+{1 \over {\sqrt n }}\right)$ [duplicate]

$$\mathop {\lim }\limits_{n \to \infty } {1 \over {\sqrt n }} \left({1 \over {\sqrt 1 }} + {1 \over {\sqrt 2 }} + {1 \over {\sqrt 3 }}+\cdots+{1 \over {\sqrt n }}\right)$$ ( Without use of integrals ). I was able to squeeze it from the bottom to ${\lim }=1$, but that's not good enough.

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## marked as duplicate by Listing, Noam D. Elkies, mrf, Davide Giraudo, Sujaan KunalanNov 28 '13 at 19:37

The tag (limit-theorems) is not a good fit for this questions, see the tag-wiki. – Martin Sleziak Nov 29 '13 at 8:01

$$\sqrt{n+1} - \sqrt{n} = \frac{1}{\sqrt{n+1} + \sqrt{n}} < \frac{1}{{\sqrt{n} + \sqrt{n}}} < \frac{1}{2\sqrt{n}}$$

add inequalities from 1 to n,

$$\sqrt{n+1} - 1 > \frac{1}{2}\left( \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots+ \frac{1}{\sqrt{n}}\right)$$

also for lower bound note that,

$$\left( \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots+ \frac{1}{\sqrt{n}}\right) > \left( \frac{1}{\sqrt{n}} + \frac{1}{\sqrt{n}} + \frac{1}{\sqrt{n}} + \cdots+ \frac{1}{\sqrt{n}}\right)= \frac{n}{\sqrt{n}}$$

so we have $$2(\sqrt{n+1} - 1) > \left( \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots+ \frac{1}{\sqrt{n}}\right) > \sqrt{n}$$

Now sandwich theorem tells us that limit is 2

EDIT:

$$\frac{1}{\sqrt{n+1} + \sqrt{n+1}} < \frac{1}{\sqrt{n}+\sqrt{n+1}} < \frac{1}{\sqrt{n}+\sqrt{n}}$$

$$\frac{1}{\sqrt{n+1} + \sqrt{n+1}}< \sqrt{n+1} - \sqrt{n} < \frac{1}{\sqrt{n}+\sqrt{n}}$$

add from 1 to n and let $$S = \left( \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots+ \frac{1}{\sqrt{n}}\right)$$,

then, $$\frac{1}{2}\left( S + \frac{1}{\sqrt{n+1}} -1\right) < \sqrt{n+1} -1 < \frac{1}{2}S$$

$$2(\sqrt{n+1} -1) <S<2(\sqrt{n+1} -1) +1 -\frac{1}{\sqrt{n+1}}$$

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Your lower bound is not good enough. It has to be $\approx 2\sqrt n$, not $\sqrt n$. – TonyK Nov 28 '13 at 19:19
Your lower bound is fine, because you don't actually have an upper bound. You don't have $$\left(\frac 1 {\sqrt1} + \frac 1 {\sqrt{2}} + ... \right) < 2(\sqrt{n+1}-1)$$ The inequality should be the other way around. – Mike Miller Nov 28 '13 at 19:21
@TonyK , improved the bounds, please see – chatur Nov 28 '13 at 19:29
@Mike, yes I got that wrong, but now should be OK – chatur Nov 28 '13 at 19:33
No, Mike is correct $-$ I didn't notice it myself, but your second inequality ("add inequalities from 1 to n") is the wrong way round. I suggest you rewrite this! – TonyK Nov 28 '13 at 19:34

We have $$(n+1)^s-n^s=n^s\left(\left(1+\frac{1}{n}\right)^s-1\right)\sim_\infty sn^{s-1}$$ so by choosing $s=\frac{1}{2}$ we find $$2\left((n+1)^{1/2}-n^{1/2}\right)\sim_\infty n^{-1/2}$$ hence we have by telescoping $$\frac{1}{\sqrt n}\sum_{k=1}^n\frac{1}{\sqrt k}\sim_\infty \frac{2}{\sqrt n}\sum_{k=1}^n \left((k+1)^{1/2}-k^{1/2}\right)=\frac{2}{\sqrt n}\left((n+1)^{1/2}-1\right)\to2$$

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Good mornin :+) – Babak S. Nov 30 '13 at 2:35
+1 and one away from a "nice answer"! – amWhy Nov 30 '13 at 16:42