Other Idea to show an inequality $\dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt n}\geq \sqrt n$

Question

$$\dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt n}\geq \sqrt n$$ I want to prove this by Induction $$n=1 \checkmark\\ n=k \to \dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt k}\geq \sqrt k\\ n=k+1 \to \dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt {k+1}}\geq \sqrt {k+1}$$ so $$\dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt k}+\dfrac{1}{\sqrt {k+1}}\geq \sqrt k+\dfrac{1}{\sqrt {k+1}}$$now we prove that $$\sqrt k+\dfrac{1}{\sqrt {k+1}} >\sqrt{k+1} \\\sqrt{k(k+1)}+1 \geq k+1 \\ k(k+1) \geq k^2 \\k+1 \geq k \checkmark$$ and the second method like below ,

and I want to know is there more Idia to show this proof ? forexample combinatorics proofs , or using integrals ,or fourier series ,....

Is there a close form for this summation ?

any help will be appreciated .

Answer 1

Combining AM-HM $$\left(a_1+a_2+...+a_n\right)\left(\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}\right)\geq n^2$$ Thus $$\left(\sqrt{1}+\sqrt{2}+...+\sqrt{n}\right)\left(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}\right)\geq n^2$$ and $$n\sqrt{n}\geq\left(\sqrt{1}+\sqrt{2}+...+\sqrt{n}\right)$$ so $$n\sqrt{n}\left(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}\right)\geq n^2$$ and $$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}\geq \sqrt{n}$$

Answer 2

$$\begin{cases}\dfrac{1}{\sqrt 1}\geq \dfrac{1}{\sqrt n}\\+\dfrac{1}{\sqrt 2}\geq \dfrac{1}{\sqrt n}\\+\dfrac{1}{\sqrt 3}\geq \dfrac{1}{\sqrt n}\\ \vdots\\+\dfrac{1}{\sqrt n}\geq \dfrac{1}{\sqrt n}\end{cases} \\\\$$

sum of left hands is $\underbrace{\dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt n}}$ sum of the right hands is $n\times \dfrac{1}{\sqrt n}$ so $$\dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt n} \geq n\dfrac{1}{\sqrt n}=\dfrac{\sqrt{n^2}}{\sqrt{n}}=\sqrt{n} \checkmark$$

Answer 3

Integrals:

$$\sum_{k=1}^n\frac1{\sqrt n}\ge\int_1^{n+1}\frac1{\sqrt x}\ dx=2\sqrt{n+1}-2$$

And it's very easy to check that

$$2\sqrt{n+1}-2\ge\sqrt n$$

for $n\ge2$.

A visuallization of this argument:

From the red lines down, that area represents a sum. From the blue line down, that represents an integral. Clearly, the integral is smaller than the sum.

Answer 4

Using arithmetic and geometric means inequality :

$\displaystyle{\frac{1}{\sqrt 1}+\frac{1}{\sqrt 2}+..+\frac{1}{\sqrt n}\ge n\,\sqrt[n]{\frac{1}{\sqrt 1\sqrt 2..\sqrt n}}\ge n\,(n!)^{-\frac1{2n}}}$

We can get an asymptotic result using Stirling formula :

$n\,(n!)^{-\frac1{2n}}\sim n\left(\sqrt{2\pi n}({\frac ne})^n\right)^{-\frac1{2n}}=\frac{\sqrt e}{\sqrt[4n]{2\pi}}\times n^{1-\frac1{4n}}\times\sqrt n=C(n)\sqrt n$ with $C(n)\to \sqrt[4]e$

Since $\sqrt[4]e\ge 1$ then $C(n)\sqrt n\ge\sqrt n$ for some $n$.

It is not as good as in the other methods, but it presents another idea.

Answer 5

By the generalized mean inequality the harmonic mean is no larger than the quadratic mean:

$$ \require{cancel} \cfrac{n}{\cfrac{1}{\sqrt{1}}+\cfrac{1}{\sqrt{2}}+\cdots+\cfrac{1}{\sqrt{n}}} \;\le\; \sqrt{\frac{(\sqrt{1})^2+(\sqrt{2})^2+\cdots+(\sqrt{n})^2}{n}} = \sqrt{\frac{\cancel{n}(n+1)}{2\,\cancel{n}}} $$

$$ \implies \quad \cfrac{1}{\sqrt{1}}+\cfrac{1}{\sqrt{2}}+\cdots+\cfrac{1}{\sqrt{n}} \;\ge\; \sqrt{\frac{2\,n^2}{n+1}} \;\ge\; \sqrt{n} $$

Answer 6

I thought it might be instructive to present a simple approach that yields a much tighter bound than requested in the OP, and that relies on nothing more than telescoping series and straightforward arithmetic. To that end, we now proceed.

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We begin with the telescoping series

$$\sum_{k=1}^{n}\left(\sqrt{k+1}-\sqrt{k}\right)=\sqrt{n+1}-1 \tag 1$$

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Inasmuch as $\sqrt{k+1}-\sqrt{k}=\frac{1}{\sqrt{k+1}+\sqrt{k}}$, we can write $(1)$ as

$$\sum_{k=1}^{n}\left(\frac{1}{\sqrt{k+1}+\sqrt{k}}\right)=\sqrt{n+1}-1 \tag 2$$

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Then, using $\sqrt{k+1}>\sqrt k$, we have the inequality

$$\sum_{k=1}^{n}\left(\frac{1}{2\sqrt{k}}\right)>\sqrt{n+1}-1$$

from which we see that

$$\bbox[5px,border:2px solid #C0A000]{\sum_{k=1}^{n}\frac{1}{\sqrt{k}}> 2(\sqrt {n+1}-1)} \tag 3$$

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Note that $(3)$ provides a much tighter bound for the summation of interest than the one requested in the OP since

$$\sum_{k=1}^n\frac{1}{\sqrt k}>2(\sqrt {n+1} -1)> \sqrt n $$

for $n\ge 2$. It's easy to see that $\sum_{k=1}^n \frac1{\sqrt k} = \sqrt n $ for $n=1$.

And we are done!

Tools Used: Telescoping Series and straightforward arithmetic