Prove that $$1 + \frac{1}{\sqrt{2}} +... +\frac{1}{\sqrt{n}} <\sqrt{n}\ .\Bigl(2n-1\Bigr)^{\frac{1}{4}} $$

My Approach :

I tried by applying Tchebychev's Inequality for two same sets of numbers;

$$1 , \frac{1}{\sqrt{2}} ,... ,\frac{1}{\sqrt{n}}$$

And got , $$\Bigl(1 + \frac{1}{\sqrt{2}} +... +\frac{1}{\sqrt{n}}\Bigr)^2 <n\Bigl(1 + \frac{1}{2} +... +\frac{1}{n}\Bigr) $$

Again I tried by applying Tchebychev's Inequality for another two same sets of numbers; $$1,\frac{1}{2},...,\frac{1}{n}$$ And got, $$\Bigl(1 + \frac{1}{2} +... +\frac{1}{n}\Bigr)^2 <n\Bigl(1 + \frac{1}{2^2} +... +\frac{1}{n^2}\Bigr)$$

With these two inequities i tried solving further more, but i couldn't. So can you please help me solving this further. And if there is some other approach for this question then please answer that way too.

Thank you.

  • $\begingroup$ Are you sure it is strict inequality, because for $n=1$ it won't hold? $\endgroup$
    – Anurag A
    Aug 8, 2018 at 6:38
  • $\begingroup$ Interestingly enough, what you have done is a proof that $1+1/\sqrt2+\ldots+1/\sqrt n<\sqrt n(2n)^{1/4}$. $\endgroup$ Aug 8, 2018 at 14:52

6 Answers 6


Assume that for any $k\geq 1$ we have $\frac{1}{k}=a_k b_k$, with $a_k$ and $b_k$ being roughly of the same magnitude and such that both $a_k$ and $b_k$ are telescopic terms. Then $$ \sum_{k=1}^{n}\sqrt{\frac{1}{k}}\leq \sqrt{\sum_{k=1}^{n}a_k}\sqrt{\sum_{k=1}^{n}b_k}, $$ which follows from the Cauchy-Schwarz inequality, is both an accurate and simple inequality. Let us see if we manage to find such $a_k$ and $b_k$. They both have to be close to $\frac{1}{\sqrt{k}}$, and on its turn $\frac{1}{\sqrt{k}}$ is pretty close to $2\sqrt{k+1/2}-2\sqrt{k-1/2}$, which is a telescopic term. With the choice

$$ a_k=2\sqrt{k+1/2}-2\sqrt{k-1/2},\quad b_k=\frac{\sqrt{k+1/2}+\sqrt{k-1/2}}{2k} $$ $b_k$ is not telescopic, but still $b_k\leq\frac{1}{\sqrt{k}}$. So by letting $S_n=\sum_{k=1}^{n}\frac{1}{\sqrt{k}}$ we get:

$$ S_n \leq \sqrt{\sqrt{4n+2}-\sqrt{2}}\sqrt{S_n} $$ and the resulting inequality $$ \sum_{k=1}^{n}\frac{1}{\sqrt{k}} = H_n^{(1/2)} \leq \sqrt{4n+2}-\sqrt{2} $$ is much sharper than $\leq \sqrt{n}(2n-1)^{1/4}$.


The strict inequality (which only holds for $n\gt1$) can be proved by induction, by showing that

$$\sqrt n\cdot(2n-1)^{1/4}+{1\over\sqrt{n+1}}\lt\sqrt{n+1}\cdot(2n+1)^{1/4}$$

for $n\ge N$ for some $N$ and then checking the base cases up to $N$.

We first rewrite the inductive inequality above as


Noting that both sides are positive for $n\ge1$, we can square to the equivalent inequality


which is certainly true if


But this reduces to $2(2n+1)^{1/4}\lt\sqrt{2n+1}$, which simplifies to $16\lt2n+1$. So the inductive inequality holds for $n\ge N=8$. And as luck would have it, the base cases for $n=2$ to $8$ have been checked in Yves Daoust's answer.


Your inequality should be not strong.

We can prove that for all natural $n\geq1$ the following inequality holds: $$\sum_{k=1}^n\frac{1}{\sqrt{k}}\leq\sqrt{n}\sqrt[4]{2n-1}.$$ Indeed, $$\sum_{k=1}^n\frac{1}{\sqrt{k}}\leq1+\int\limits_1^{n}\frac{1}{\sqrt x}dx=2\sqrt{n}-1.$$ Thus, it's enough to prove that $$2\sqrt{n}-1\leq\sqrt{n}\sqrt[4]{2n-1}.$$ Let $n=(x+1)^2,$ where $x\geq0$.

Thus, we need to prove that $$2x+1\leq (x+1)\sqrt[4]{2x^2+4x+1}$$ or $$(x+1)^4(2x^2+4x+1)\geq(2x+1)^4$$ or $$x^3(2x^3+12x^2+13x+4)\geq0,$$ which is obvious.


  • $\begingroup$ Without much computation, one can add that the ratio RHS/LHS is asymptotic to $n^{1/4}$ and is certainly $\ge1$ as of some $n$. $\endgroup$
    – user65203
    Aug 8, 2018 at 7:06
  • $\begingroup$ Seems to me that the first inequality should be reversed. $\endgroup$ Aug 8, 2018 at 7:06
  • $\begingroup$ Obviously the original inequality does not hold for $n=1$. $\endgroup$ Aug 8, 2018 at 7:25
  • $\begingroup$ @uniquesolution I fixed my post again. See now. $\endgroup$ Aug 8, 2018 at 8:30

$$\frac1{\sqrt{\lceil x\rceil}}\le\frac1{\sqrt x}$$ and integrating from $0$ ("excluded") to $n$, $$\sum_{k=1}^n\frac1{\sqrt k}<\int_0^n\frac{dx}{\sqrt x}=2\sqrt n.$$

This leads us to

$$2\sqrt n<\sqrt n\sqrt[4]{2n-1}.$$

This inequality is certainly true for $n>8$, and we can check the remaining values explicitly.

$$1 \le 1 \\ 1.70710678119 < 1.8612097182 \\ 2.28445705038 < 2.59002006411 \\ 2.78445705038 < 3.2531531234 \\ 3.23167064588 < 3.87298334621 \\ 3.63991893634 < 4.46091344257 \\ 4.01788340935 < 5.02382911018 \\ 4.37143679994 < 5.56631536743 \\ 4.70477013328 < 6.09162955461 \\\cdots$$

  • $\begingroup$ I think my estimation is better. See please my solution. $\endgroup$ Aug 8, 2018 at 7:19
  • 1
    $\begingroup$ @uniquesolution: we know that, why do you say it ? $\endgroup$
    – user65203
    Aug 8, 2018 at 7:25
  • $\begingroup$ @MichaelRozenberg: right. This is because in the beginning of the curve, the integral is a poor approximation. We could yet improve with more terms, like $1+1/\sqrt2+\int_2^n\cdots$. $\endgroup$
    – user65203
    Aug 8, 2018 at 7:26
  • 1
    $\begingroup$ @uniquesolution: the way I wrote, we know that inequality holds for $n>1$ and equality for $n=1$. $\endgroup$
    – user65203
    Aug 8, 2018 at 7:28

Here is a slightly different start than that of @michael-rozenberg, avoiding integrals in case one would like to:

Start by writing (assuming $n>1$) $$ \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{n}}< \frac{1}{\sqrt{1}}+\frac{2}{\sqrt{2}+\sqrt{1}}+\cdots+\frac{2}{\sqrt{n}+\sqrt{n-1}}. $$ Multiplying by conjugates you will find that the sum on the right-hand side equals $$ 1+2\bigl((\sqrt{2}-\sqrt{1})+\cdots+(\sqrt{n}-\sqrt{n-1})\bigr)=1+2(\sqrt{n}-1). $$ Thus, your sum is less than $$ 2\sqrt{n}-1. $$ Then continue in the same manner as in the nice solution by @michael-rozenberg.


Using Cauchy Schwartz inequality with

  • $u=(1,1,...,1)$

  • $v=\left(1 , \frac{1}{\sqrt{2}} ,... ,\frac{1}{\sqrt{n}}\right)$

we have

$$\vec u \cdot \vec v\le |\vec u||\vec v| \implies 1 + \frac{1}{\sqrt{2}} +... +\frac{1}{\sqrt{n}} \le\sqrt{n}\ \cdot\left(\sqrt{1 + \frac{1}{2} +... +\frac{1}{n}}\right)$$

and thus it suffices to prove that

$$\sqrt{n}\ \cdot\left(\sqrt{1 + \frac{1}{2} +... +\frac{1}{n}}\right)<\sqrt{n}\ .\left(2n-1\right)^{\frac{1}{4}}$$

that is

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

which is true for $n=2$ and it reduces to prove by induction


which is true indeed









Therefore the given inequality holds for $n\ge 2$, for $n=1$ we can check directly that equality holds.

  • $\begingroup$ So $u\cdot v$ is the l.h.s of the OP's inequality. C-S gives $\|u\|\cdot \|v\|$ on the r.h.s, which is equal to $\sqrt{n}\left(\sum_{k=1}^n(1/k)\right)^{1/2}$, so by the inequality you say that can easily be proven by induction, the result on the r.h.s is $\sqrt{n}\sqrt{2n-1}$. How do you get to $(2n-1)^{1/4}$? $\endgroup$ Aug 8, 2018 at 7:11
  • $\begingroup$ One needs to show $1 + \frac{1}{2} +... +\frac{1}{n}<\sqrt{2n-1}$ rather than $1 + \frac{1}{2} +... +\frac{1}{n}<2n-1$. $\endgroup$ Aug 8, 2018 at 7:13
  • $\begingroup$ Ops sorry I did it mentally and lost a square factot! $\endgroup$
    – user
    Aug 8, 2018 at 7:14

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