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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!

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    $\begingroup$ Please do not ask multiple questions on a single post $\endgroup$
    – newbie105
    Jul 7, 2021 at 17:03
  • $\begingroup$ I will take care of this. $\endgroup$
    – JINENDRA
    Jul 7, 2021 at 17:11
  • $\begingroup$ Note $\sqrt 2= (2!)^{1/2}.$ $\endgroup$
    – zhw.
    Jul 7, 2021 at 18:19

4 Answers 4

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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 $1<x<n$, 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.

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  • $\begingroup$ thanks.. can you please give me hints for the first one? $\endgroup$
    – JINENDRA
    Jul 7, 2021 at 16:56
  • $\begingroup$ @JINENDRA Oh! There was that as well? I didn't notice. Wait. $\endgroup$ Jul 7, 2021 at 16:59
  • $\begingroup$ @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. $\endgroup$ Jul 7, 2021 at 17:00
  • $\begingroup$ @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. $\endgroup$ Jul 7, 2021 at 17:03
  • $\begingroup$ 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. $\endgroup$
    – René Gy
    Jul 7, 2021 at 21:05
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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 $x \ge 1$.

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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.

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This is my final answer.

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} $

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