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 '18 at 6:38
  • $\begingroup$ Interestingly enough, what you have done is a proof that $1+1/\sqrt2+\ldots+1/\sqrt n<\sqrt n(2n)^{1/4}$. $\endgroup$ – Cave Johnson Aug 8 '18 at 14:52

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 $n$. $\endgroup$ – Yves Daoust Aug 8 '18 at 7:06
  • $\begingroup$ Seems to me that the first inequality should be reversed. $\endgroup$ – uniquesolution Aug 8 '18 at 7:06
  • $\begingroup$ Obviously the original inequality does not hold for $n=1$. $\endgroup$ – uniquesolution Aug 8 '18 at 7:25
  • $\begingroup$ @uniquesolution I fixed my post again. See now. $\endgroup$ – Michael Rozenberg Aug 8 '18 at 8:30

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 $\leq \sqrt{n}(2n-1)^{1/4}$.


$$\frac1{\sqrt{\lceil x\rceil}}\le\frac1{\sqrt x}$$ 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. $\endgroup$ – Michael Rozenberg Aug 8 '18 at 7:19
  • 1
    $\begingroup$ @uniquesolution: we know that, why do you say it ? $\endgroup$ – Yves Daoust Aug 8 '18 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 $1+1/\sqrt2+\int_2^n\cdots$. $\endgroup$ – Yves Daoust Aug 8 '18 at 7:26
  • 1
    $\begingroup$ @uniquesolution: the way I wrote, we know that inequality holds for $n>1$ and equality for $n=1$. $\endgroup$ – Yves Daoust Aug 8 '18 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 $(2n-1)^{1/4}$? $\endgroup$ – uniquesolution Aug 8 '18 at 7:11
  • $\begingroup$ One needs to show $1 + \frac{1}{2} +... +\frac{1}{n}<\sqrt{2n-1}$ rather than $1 + \frac{1}{2} +... +\frac{1}{n}<2n-1$. $\endgroup$ – Jens Schwaiger Aug 8 '18 at 7:13
  • $\begingroup$ Ops sorry I did it mentally and lost a square factot! $\endgroup$ – gimusi Aug 8 '18 at 7:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.