Limit of the sum of a symmetric polynomial evaluated at sequence terms

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Limit of the sum of a symmetric polynomial evaluated at sequence terms

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Source: 2018 Osaka City University entrance exam

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#1 h Apr 29, 2025, 1:27 AM

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The real sequence $\{ a_n \}_{n=1,2,3,\ldots}$ satisfies $0 \le a_n \le 1 \, (\forall n \in \mathbb{N} )$ and $\lim_{n\to\infty} a_n = 1$ . Find
$\lim_{n\to\infty} n \left( \;1 \; +\; \sum_{k=1}^n \frac{ (-1)^k}{k+1} \sum_{1 \le j_1 < j_2 < \cdots < j_k \le n} a_{j_1} a_{j_2} \cdots a_{j_k} \right).$

This post has been edited 1 time. Last edited by tom-nowy, Apr 30, 2025, 1:21 PM

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#3 h Apr 30, 2025, 1:10 PM • 1 Y

Y by tom-nowy

In the second sum, are you sure it's $j_1 \le j_2 \le \cdots \le j_k$ and not $j_1 < j_2 < \cdots < j_k ?$ I deleted my answer because I thought it was $j_1 < j_2 < \cdots < j_k .$

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#5 h Apr 30, 2025, 1:23 PM

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Sorry, I got it wrong. $j_1 < j_2 < \cdots < j_k$ is correct. I've edited it.

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#6 h Apr 30, 2025, 1:43 PM • 1 Y

Y by tom-nowy

Ok, thanks. So here's my answer again.

The limit is $1.$ First, see that
$b_n:=1+\sum_{k=1}^n \frac{ (-1)^k}{k+1}\sum_{1 \le j_1 < j_2 < \cdots < j_k \le n}a_{j_1} a_{j_2} \cdots a_{j_k}=\int_0^1(1-a_1x) \cdots (1-a_nx) \ dx.$ Now, since $a_k \in [0,1]$ for all $k,$
$(1-a_1x) \cdots (1-a_nx) \ge (1-x)^n,$ and, since $1+t \le e^t$ for all $t,$
$(1-a_1x) \cdots (1-a_nx) \le e^{-(a_1+ \cdots + a_n)x}.$ Thus
$\frac{n}{n+1}=n\int_0^1(1-x)^ndx \le nb_n \le n\int_0^1e^{-(a_1+ \cdots + a_n)x}dx=\frac{n}{a_1+ \cdots + a_n}\left(1-e^{-(a_1+ \cdots +a_n)} \right)$ and so $\lim_{n\to\infty}nb_n=1$ because
$\lim_{n\to\infty}(a_1+\cdots +a_n)=\infty, \ \ \ \ \lim_{n\to\infty}\frac{a_1+ \cdots +a_n}{n}=\lim_{n\to\infty}a_n=1.$
Question. If we replace the condition $\lim_{n\to\infty}a_n=1$ with $\lim_{n\to\infty}a_n=a \in(0,1],$ then we'll have $\lim_{n\to\infty}nb_n=\frac{1}{a} ?$

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