The sequence $A_n=\prod_{k=1}^n\left(1+\frac{k}{n^2}\right)$ is decreasing

Question

Let $A$ be the sequence of real numbers defined by :

$$\forall n\in\mathbb{N}^\star,\,A_n=\prod_{k=1}^n\left(1+\frac{k}{n^2}\right)$$

I know how to prove that this sequence converges to $\sqrt e$, using the following inequalities :

$$\forall t>0,\,t-\frac{t^2}2\leqslant\ln(1+t)\leqslant t$$

I found numerical evidence that $(A_n)$ is decreasing, but wasn't able to prove it. Any help will be appreciated.

Answer 1

A few observations to begin with

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Pr1. $1+x\leq e^x$ and $1-x\leq e^{-x}$ for $\forall x$.

These inequalities are well known.

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Pr2. $$\sum\limits_{k=1}^n \frac{1}{n^2+k} < \int\limits_{n^2+1}^{n^2+n+1}\frac{1}{x}dx=\log{\left(1+\frac{n}{n^2+1}\right)}$$

Easy to show using Riemann_sum.

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Pr3. Function $f(x)=\log{\left(1+\frac{x}{x^2+1}\right)}-\frac{2x^2-1}{x^2(2x+1)} \leq 0$ for $\forall x\geq2$.

Easy to check from $f(2)\approx -0.013527763 <0$ and $\lim\limits_{x\to\ +\infty}f(x)=0$. Then $$f'(x)= \frac{(5x^5-9x^3-10x^2-8x-2)}{x^3(2x+1)^2(x^2+1)(x^2+x+1)} >0, \forall x\geq2$$ which means $f(x)$ is ascending for $\forall x\geq2$, i.e. it ascends to $0$ from $f(2)<0$. If we assume $\exists x_0>2: f(x_0)>0$, then (because $f$ is ascending) $f(x)\geq f(x_0)>0$ for $\forall x\geq x_0>2$. This contradicts the fact that $\lim\limits_{x\to\ +\infty}f(x)=0$.

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Pr4. $\frac{n^2}{(n+1)^2}+\frac{n^2}{(n+1)^2}\cdot\frac{2n+1}{n^2+k} = 1-\frac{k(2n+1)}{(n^2+k)(n+1)^2}$

From $$\frac{n^2}{(n+1)^2}+\frac{n^2}{(n+1)^2}\cdot\frac{2n+1}{n^2+k} - 1 =\\ \frac{n^2(n^2+k)+n^2(2n+1)-(n^2+k)(n+1)^2}{(n^2+k)(n+1)^2}= -\frac{k(2n+1)}{(n^2+k)(n+1)^2}$$

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Now, let's rewrite $A_n$ as $$A_n=\prod\limits_{k=1}^n\left(1+\frac{k}{n^2}\right)= \frac{1}{n^{2n}}\prod\limits_{k=1}^n(n^2+k)$$

And check $$\frac{A_{n+1}}{A_n}=\frac{\frac{1}{(n+1)^{2(n+1)}}\prod\limits_{k=1}^{n+1}((n+1)^2+k)}{\frac{1}{n^{2n}}\prod\limits_{k=1}^n(n^2+k)}=\\ \frac{n^{2n}}{(n+1)^{2(n+1)}}\cdot((n+1)^2+n+1)\cdot\prod\limits_{k=1}^n\frac{(n+1)^2+k}{n^2+k}=\\ \left(\frac{n}{n+1}\right)^{2n}\cdot\left(1+\frac{1}{n+1}\right)\cdot\prod\limits_{k=1}^n\left(1+\frac{2n+1}{n^2+k}\right)=\\ \left(1+\frac{1}{n+1}\right)\cdot\prod\limits_{k=1}^n\left(\frac{n^2}{(n+1)^2}+\frac{n^2}{(n+1)^2}\cdot\frac{2n+1}{n^2+k}\right)\overset{\color{red}{Pr4}}{=}\\ \left(1+\frac{1}{n+1}\right)\cdot\prod\limits_{k=1}^n\left(1-\frac{k(2n+1)}{(n^2+k)(n+1)^2}\right) \overset{\color{red}{Pr1}}{\leq} $$ $$e^{\frac{1}{n+1}} \cdot e^{-\sum\limits_{k=1}^n \frac{k(2n+1)}{(n^2+k)(n+1)^2}}= e^{\color{blue}{\frac{1}{n+1}-\frac{2n+1}{(n+1)^2}\left(\sum\limits_{k=1}^n \frac{k}{n^2+k}\right)}} \tag{1}$$

Let's look at $$\color{blue}{\frac{1}{n+1} - \frac{2n+1}{(n+1)^2}\left(\sum\limits_{k=1}^n \frac{k}{n^2+k}\right)} < 0 \iff \\ n+1 - (2n+1)\left(\sum\limits_{k=1}^n \frac{k}{n^2+k}\right) <0 \iff \\ \sum\limits_{k=1}^n \frac{k}{n^2+k} > \frac{n+1}{2n+1}=1-\frac{n}{2n+1} \iff \\ \sum\limits_{k=1}^n \left(1-\frac{n^2}{n^2+k}\right) > 1-\frac{n}{2n+1} \iff \\ -\sum\limits_{k=1}^n \frac{n^2}{n^2+k} > 1-\frac{n}{2n+1}-n=-\frac{2n^2-1}{2n+1} \iff $$ $$\sum\limits_{k=1}^n \frac{1}{n^2+k} < \frac{2n^2-1}{n^2(2n+1)} \tag{2}$$ which is true for large enough $n$ since $$\sum\limits_{k=1}^n \frac{1}{n^2+k} \overset{\color{red}{Pr2}}{<} \log{\left(1+\frac{n}{n^2+1}\right)} \overset{\color{red}{Pr3}}{\leq} \frac{2n^2-1}{n^2(2n+1)}$$

Returning back to $(1)$ $$\frac{A_{n+1}}{A_n} < e^0=1$$ for large enough $n$.

Answer 2

Fun with Stirling.

I get $\ln(A_n) =\frac12+\frac1{3n}+O(\frac1{n^2}) $.

Numerically this checks.

Here's how.

$\begin{array}\\ A_n &=\prod_{k=1}^n\left(1+\dfrac{k}{n^2}\right)\\ B_n &=\ln(A_n)\\ &=\sum_{k=1}^n\ln\left(1+\dfrac{k}{n^2}\right)\\ &=\sum_{k=1}^n\ln\left(\dfrac{n^2+k}{n^2}\right)\\ &=\sum_{k=1}^n\ln(n^2+k)-\sum_{k=1}^n\ln(n^2)\\ &=\sum_{k=n^2+1}^{n^2+n}\ln(k)-n\ln(n^2)\\ &=\sum_{k=1}^{n^2+n}\ln(k)-\sum_{k=1}^{n^2}\ln(k)-2n\ln(n)\\ &=\ln((n^2+n)!)-\ln((n^2)!)-2n\ln(n)\\ &\approx\frac12\ln(n^2+n)+(n^2+n)(\ln(n^2+n)-1)+O(1/n^2))-(\frac12\ln(n^2)+n^2(\ln(n^2)-1)+O(1/n^2))-2n\ln(n)\\ &=\frac12\ln(1+1/n)+(n^2+n)(\ln(n^2+n)-1))-(n^2(\ln(n^2)-1)+O(1/n^2))-2n\ln(n)\\ &=\frac12\ln(1+1/n)+n^2(\ln(n^2+n)-1))-(n^2(\ln(n^2)-1)+n(\ln(n^2+n)-1))+O(1/n^2))-2n\ln(n)\\ &=\frac12\ln(1+1/n)+n^2\ln(1+1/n)+n(\ln(n^2+n)-1))+O(1/n^2))-2n\ln(n)\\ &=\frac12\ln(1+1/n)+n^2\ln(1+1/n)+n(2\ln(n)+\ln(1+1/n)-1))+O(1/n^2))-2n\ln(n)\\ &=\frac12\ln(1+1/n)+n^2\ln(1+1/n)+n(\ln(1+1/n)-1))+O(1/n^2))\\ &=(n^2+n+\frac12)\ln(1+1/n)-n+O(1/n^2))\\ &=(n^2+n+\frac12)(\frac1{n}-\frac1{2n^2}+\frac1{3n^3}+O(\frac1{n^4}))-n+O(1/n^2))\\ &=(n+\frac12+\frac1{3n}+O(\frac1{n^2}))-n+O(1/n^2))\\ &=\frac12+\frac1{3n}+O(\frac1{n^2})\\ \end{array} $

Answer 3

It is easy to verify $A_1 > A_2 > A_3 > A_4$. It suffices to prove that $A_n > A_{n+1}$ for all $n\ge 4$. It suffices to prove that, for all $n\ge 4$, $$\sum_{k=1}^n \ln (1 + k/n^2) > \sum_{k=1}^{n+1} \ln (1 + k/(n+1)^2)$$ or $$\sum_{k=1}^n \ln \frac{1 + k/n^2}{1 + k/(n+1)^2} > \ln (1 + 1/(n+1))$$ or $$\sum_{k=1}^n \ln\left(1 + \frac{k(2n+1)}{n^2(n^2+k+2n+1)} \right) > \ln (1 + 1/(n+1)).$$ By using $\ln (1+x) \ge \frac{x}{1+x}$ for $x > 0$, we have \begin{align} \ln\left(1 + \frac{k(2n+1)}{n^2(n^2+k+2n+1)} \right) &\ge \frac{k(2n+1)}{(n^2+k)(n+1)^2}\\ &= \frac{k(2n+1)}{n^2(n+1)^2}\, \frac{1}{1 + k/n^2}\\ &\ge \frac{k(2n+1)}{n^2(n+1)^2}(1 - k/n^2). \end{align} Also, by using $\ln(1+x) < \frac{x^2+6x}{6+4x}$ for $x > 0$, we have $$\ln (1 + 1/(n+1)) < \frac{7+6n}{2(3n+5)(n+1)}.$$ Thus, it suffices to prove that, for all $n\ge 4$, $$\sum_{k=1}^n \frac{k(2n+1)}{n^2(n+1)^2}(1 - k/n^2) > \frac{7+6n}{2(3n+5)(n+1)}$$ or $$\frac{(n-1)(2n+1)(3n+1)}{6n^3(n+1)} > \frac{7+6n}{2(3n+5)(n+1)}$$ or $$\frac{6n^3-17n^2-23n-5}{6n^3(n+1)(3n+5)} > 0.$$ It is true. We are done.