# $\sqrt n < n!^\frac1n\ \forall \ n$

Prove the following:

#1} $$\frac12 < \frac1{3n+1} + \frac1{3n+2} + … + \frac1{5n+1} < \frac23$$

#2} $$\sqrt n < n!^\frac1n \ \forall\ n>2$$

Attempt (i) to #1}

Since $$\frac{2n+1}{5n+1} < \frac1{3n+1} + \frac1{3n+2} + … + \frac1{5n+1} < \frac{2n+1}{3n+1}$$

$$\implies \frac13 < \frac1{3n+1} + \frac1{3n+2} + … + \frac1{5n+1} < 1.$$

Attempt (ii) to #1} Since A.M.≥ H.M.

By some calculations, I got $$\frac1{3n+1} + \frac1{3n+2} + … + \frac1{5n+1} > \frac25$$

But, from either of the attempts nothing has proved.

And for #2} i have no idea how to get that. Had tried to use A.M.≥G.M.≥H.M.

Any help is really appreciated!

• Please do not ask multiple questions on a single post Jul 7, 2021 at 17:03
• I will take care of this. Jul 7, 2021 at 17:11
• Note $\sqrt 2= (2!)^{1/2}.$
– zhw.
Jul 7, 2021 at 18:19

I will show that $$n^n<(n!)^2$$. Note that that's not true for all $$n\in \mathbb N$$ as mentioned in the question. That's only true for $$n>2$$.

Now, note that for any integer $$x$$ which satisfies $$1, we have $$(x-1)>0$$ and $$(x-n)<0$$, which together gives $$(x-1)(x-n)<0$$ which implies $$n< x\cdot (n-x+1)$$. Also, for $$x=1$$ or $$x=n$$, we have $$n\leq x\cdot (n-x+1)$$.

Put $$x=1,2,3,\dots ,n$$ to get \begin{align*} &n\leq 1\cdot n\\ &n<2\cdot (n-1)\\ &n<3\cdot (n-2)\\ &\dots\\ &\dots\\ &\dots\\ &n\leq n\cdot 1 \end{align*} Multiply these equations to get $$n^n<(n!)^2$$

This completes the proof.

• thanks.. can you please give me hints for the first one? Jul 7, 2021 at 16:56
• @JINENDRA Oh! There was that as well? I didn't notice. Wait. Jul 7, 2021 at 16:59
• @JINENDRA If the source of these problems is what I think it is, then ctanujit.org/uploads/2/5/3/9/25393293/… might help you. Jul 7, 2021 at 17:00
• @JINENDRA About the first problem, I'm not quite sure, but try the standard approach of integrating $f(x)=\frac 1 x$ from $x=3n+1$ to $x=5n+1$. I think, that will help. Jul 7, 2021 at 17:03
• It might be better to write $n \le x \cdot (n-x+1)$ unconditionally for all $x$ such that $1 \le x \le n$ and that as soon as $n \gt 2$, there exists $y$ such that $1 \lt y \lt n$, which is enough to obtain the strict inequality after the multiplications. Jul 7, 2021 at 21:05

Here's a proof by induction of the second inequality.

Suppose $$n^n < (n!)^2$$ and $$(n+1)^{n+1} \ge ((n+1)!)^2$$.

Then

$$\begin{array}\\ (n+1)^{n+1} &\ge ((n+1)!)^2\\ &=(n!)^2(n+1)^2\\ &\gt n^n(n+1)^2\\ \text{so}\\ (n+1)^{n-1} &\gt n^n\\ \text{or}\\ (1+\frac1{n})^n &\gt n\\ \end{array}$$

which is false for $$n \ge 3$$ since $$(1+\frac1{n})^n \lt e \lt 3$$.

Therefore $$n^n < (n!)^2$$ and $$n \ge 3$$ implies $$(n+1)^{n+1} \lt ((n+1)!)^2$$.

Since $$3^3 =27 \lt 36 =(3!)^2$$, $$n^n < (n!)^2$$ for $$n \ge 3$$.

You can also argue that $$(n+1)^{n-1} \gt n^n$$ implies $$(n+1)^{1/n} \gt n^{1/(n-1)}$$ which is false for $$n \ge 2$$ since $$f(x) =(x+1)^{1/x}$$ is decreasing for $$x \ge 1$$.

I can show that the first sum is between $$\dfrac{40}{81}$$ and $$\dfrac{44}{81}+\dfrac{25}{81n} \lt\dfrac23$$, so this is not quite the lower bound.

$$\begin{array}\\ s(n) &=\sum_{k=3n+1}^{5n+1} \dfrac1{k}\\ &=\sum_{k=1}^{2n+1} \dfrac1{3n+k}\\ &=\dfrac1{3n}\sum_{k=1}^{2n+1} \dfrac1{1+k/(3n)}\\ &\lt\dfrac1{3n}\sum_{k=1}^{2n+1} (1-k/(3n)+k^2/(3n)^2) \qquad\dfrac1{1+x} \lt 1-x+x^2\\ &=\dfrac{2n+1}{3n}-\dfrac1{3n}\sum_{k=1}^{2n+1}\dfrac{k}{3n}+\dfrac1{3n}\sum_{k=1}^{2n+1}\dfrac{k^2}{9n^2}\\ &=\dfrac23+\dfrac1{3n}-\dfrac1{9n^2}\dfrac{(2n+1)(2n+2)}{2}+\dfrac1{27n^3}\dfrac{(2n+1)(2n+2)(4n+3)}{6}\\ &=\dfrac23+\dfrac1{3n}-\dfrac1{9n^2}(2n+1)(n+1)+\dfrac1{81n^3}(2n+1)(n+1)(4n+3)\\ &=\dfrac{44}{81}+\dfrac1{27 n^3} + \dfrac{4}{81 n^2} + \dfrac{2}{9 n}\\ &\lt\dfrac{44}{81}+\dfrac{25}{81n} \qquad\text{for } n \ge 2\\ &\lt \dfrac23 \qquad\text{for } n \ge 3 \quad(\dfrac23-\dfrac{44}{81}=\dfrac{10}{81})\\ \end{array}$$

$$\begin{array}\\ s(n) &=\sum_{k=3n+1}^{5n+1} \dfrac1{k}\\ &=\sum_{k=1}^{2n+1} \dfrac1{3n+k}\\ &=\dfrac1{3n}\sum_{k=1}^{2n+1} \dfrac1{1+k/(3n)}\\ &\gt\dfrac1{3n}\sum_{k=1}^{2n+1} (1-k/(3n)+k^2/(3n)^2-k^3/(3n)^3) \qquad\dfrac1{1+x} \gt 1-x+x^2-x^3\\ &=\dfrac{2n+1}{3n}-\dfrac1{3n}\sum_{k=1}^{2n+1}\dfrac{k}{3n}+\dfrac1{3n}\sum_{k=1}^{2n+1}\dfrac{k^2}{9n^2}-\dfrac1{3n}\sum_{k=1}^{2n+1}\dfrac{k^3}{27n^3}\\ &=\dfrac23+\dfrac1{3n}-\dfrac1{9n^2}\dfrac{(2n+1)(2n+2)}{2}+\dfrac1{27n^3}\dfrac{(2n+1)(2n+2)(4n+3)}{6}-\dfrac1{81n^4}\dfrac{(2n+1)^2(2n+2)^2}{4}\\ &=\dfrac23+\dfrac1{3n}-\dfrac1{9n^2}(2n+1)(n+1)+\dfrac1{81n^3}(2n+1)(n+1)(4n+3)-\dfrac1{81n^4}(2n+1)^2(n+1)^2\\ &=\dfrac{40}{81}-\dfrac1{81 n^4} -\dfrac1{27 n^3} - \dfrac1{9 n^2} + \dfrac{2}{27 n} \\ \end{array}$$

So not quite $$\dfrac12$$.

If I used $$\dfrac1{1+x} \gt 1-x+x^2-x^3+x^4-x^5$$, the bound might be improved.

The sum is $$\ln(5/3)+\dfrac{3}{5n}+O(n^{-2}) \approx 0.5108$$
In general, $$\sum_{k=un+1}^{vn+1} \dfrac1{k} =\ln(v/u)+\dfrac{u}{vn}+O(n^{-2})$$.
$$\begin{array}\\ s(n) &=\sum_{k=3n+1}^{5n+1} \dfrac1{k}\\ &=\sum_{k=1}^{2n+1} \dfrac1{3n+k}\\ &=\dfrac1{3n}\sum_{k=1}^{2n+1} \dfrac1{1+k/(3n)}\\ &=\dfrac1{3n}\sum_{k=1}^{2n+1} \sum_{m=0}^{\infty}\dfrac{(-1)^mk^m}{3^mn^m}\\ &=\dfrac1{3n}\sum_{m=0}^{\infty}\dfrac{(-1)^m}{3^mn^m}\sum_{k=1}^{2n+1} k^m\\ &=\sum_{m=0}^{\infty}\dfrac{(-1)^m}{3^{m+1}n^{m+1}} \left(\dfrac{(2n+1)^{m+1}}{m+1}+\dfrac{(2n+1)^m}{2}+O(n^{m-1}) \right)\\ &=\sum_{m=0}^{\infty}\dfrac{(-1)^m}{3^{m+1}n^{m+1}} \left(\dfrac{(2n)^{m+1}+(m+1)(2n)^{m}+O(n^{m-1})}{m+1}+\dfrac{(2n)^m+O(n^{m-1})}{2}+O(n^{m-1}) \right)\\ &=\sum_{m=0}^{\infty}\dfrac{(-1)^m}{3^{m+1}} \left(\dfrac{(2)^{m+1}+(m+1)(2^m/n+O(n^{-2}))}{m+1}+\dfrac{2^m/n+O(n^{-2})}{2}+O(n^{-2}) \right)\\ &=\sum_{m=0}^{\infty}\dfrac{(-1)^m}{3^{m+1}} \left(\dfrac{2^{m+1}}{m+1}+\dfrac1{n}\left( 2^m+\dfrac{2^m}{2}\right)+O(n^{-2}) \right)\\ &=\sum_{m=0}^{\infty}\dfrac{(-1)^m(2/3)^{m+1}}{m+1}+\dfrac1{n}\sum_{m=0}^{\infty}(-1)^{m}(2/3)^m+O(n^{-2}) \\ &=\ln(5/3)+\dfrac{3}{5n}+O(n^{-2})\\ &\approx 0.5108\\ \end{array}$$