$$\dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt n}\geq \sqrt n$$ 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 .


6 Answers 6


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

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 $$\dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt n} \geq n\dfrac{1}{\sqrt n}=\dfrac{\sqrt{n^2}}{\sqrt{n}}=\sqrt{n} \checkmark$$

  • $\begingroup$ For an alternative interpretation of the same concept, divide by $n$ and observe that the LHS is the mean value and the RHS is the minimum value of the function $f(x)=1/\sqrt{x}$ on $[n]$ $\endgroup$ Commented Apr 4, 2017 at 18:30
  • 1
    $\begingroup$ Great answer! +1 $\endgroup$
    – Klangen
    Commented Jan 25, 2020 at 18:14


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

And it's very easy to check that

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

for $n\ge2$.

A visuallization of this argument:

enter image description here

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.

  • $\begingroup$ :Is there a visual explanation for that proof ? $\endgroup$
    – Khosrotash
    Commented Feb 18, 2017 at 0:16
  • $\begingroup$ Of course. Give me a few moments... $\endgroup$ Commented Feb 18, 2017 at 0:16
  • $\begingroup$ thanks a lot .it was interesting $\endgroup$
    – Khosrotash
    Commented Feb 18, 2017 at 0:18
  • $\begingroup$ No problem. The few moments are over :-) $\endgroup$ Commented Feb 18, 2017 at 0:20
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    $\begingroup$ Interestingly we can obtain the same bound without appealing to integrals. I've posted a solution herein and would like to hear your thoughts. ;-)) -Mark $\endgroup$
    – Mark Viola
    Commented Feb 18, 2017 at 6:05

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.

We begin with the telescoping series

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

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

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

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


from which we see that

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

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

  • $\begingroup$ (+1) I find answers such as this very appealing. What made you think of telescoping? $\endgroup$
    – Plopperzz
    Commented Feb 18, 2017 at 7:01
  • $\begingroup$ @Plopperzz Thank you! Much appreciative. If we use the first term of the EMSF it is $2(\sqrt n - 1)$. This looks like the result of a telescoping series. -Mark $\endgroup$
    – Mark Viola
    Commented Feb 18, 2017 at 7:25
  • $\begingroup$ (+1) very nicely done. Very nice use of the difference of two squares. $\endgroup$ Commented Feb 18, 2017 at 11:23
  • $\begingroup$ (+1) Nice. though too much yellow for my taste. A little typo at the end, should be $\ge \sqrt n$. $\endgroup$
    – zwim
    Commented Feb 18, 2017 at 13:47
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    $\begingroup$ "I really appreciate it." And now we see the hazard from using a "not-so-smart" phone to post here. $\endgroup$
    – Mark Viola
    Commented Feb 18, 2017 at 16:28

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

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

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

  • 1
    $\begingroup$ Nice one, I compared $M_0$ and $M_1$ while comparing $M_{-1}$ and $M_2$ was bringing the result directly. $\endgroup$
    – zwim
    Commented Feb 18, 2017 at 2:53

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 $$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}\geq \sqrt{n}$$

  • $\begingroup$ That's kind of perverse to use $\sqrt k\le\sqrt n$ to get $n\sqrt n$ and not use it for $\frac{1}{\sqrt k}\ge\frac{1}{\sqrt n}$ as in Khosrotash solution. $\endgroup$
    – zwim
    Commented Feb 20, 2017 at 0:21
  • $\begingroup$ @zwim as long as it works $\endgroup$
    – rtybase
    Commented Feb 20, 2017 at 0:29

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.

  • 1
    $\begingroup$ One may acquire exact inequalities by noting that:$$n!=\exp\left[\sum\ln(k)\right]\le\exp\left[\int\ln(x)\ dx\right]=e\sqrt n\left(\frac ne\right)^n$$ $\endgroup$ Commented Feb 18, 2017 at 1:02
  • $\begingroup$ I'm not sure what you mean, but since the $n!$ is raised to a negative power, we must find an upper bound, which flips to a lower bound, which then provides what we need quite nicely. $\endgroup$ Commented Feb 18, 2017 at 1:14
  • $\begingroup$ Yes, nice one :p $\endgroup$
    – zwim
    Commented Feb 18, 2017 at 1:14

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