How to prove $1 + \frac{1}{\sqrt{2}} +... +\frac{1}{\sqrt{n}} <\sqrt{n}\ .\left(2n-1\right)^{1/4} $

Question

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

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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.

Answer 1

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

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

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

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

which is certainly true if

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

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.

Answer 2

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

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

which is true indeed

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

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

$$4n-\frac{1}{(n+1)^2}>2\sqrt{4n^2-1}$$

$$16n^2+\frac{1}{(n+1)^4}-\frac{8n}{(n+1)^2}>16n^2-4$$

$$\frac{1}{(n+1)^4}-\frac{8n}{(n+1)^2}+4>0$$

$$4(n+1)^4-8n(n+1)^2+1>0$$

$$4(n^4+4n^3+6n^2+4n+1)-8n(n^2+2n+1)+1>0$$

$$4n^4+8n^3+8n^2+8n+5>0$$

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

Answer 3

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.

Done!

Answer 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.

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

Answer 5

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.

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