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office 某人問我以下一條從 probability 書抽出來的 exercise：

Problem. Suppose that $\{c_{jn}\in \R: j\leq n, n=1,2,3,\dots \}$ satisfies $\dis \lim_{n\to \infty}\sum_{j=1}^n c_{jn}=\lambda$ and $\dis \limn \max_{1\leq j\leq n}|c_{jn}|=0$, prove that $$\limn \prod_{j=1}^n  (1+c_{jn})=e^\lambda=\exp \brac{ \lim_{n\to \infty}\sum_{j=1}^n c_{jn}}.$$  

不難的，可作為不錯的 Math2033 exercise。 這 identiy 告訴我們只要知道 $\limn\sum_{j=1}^n c_{jn}$ 就可以計算相關的 infinite product。當 $c_{jn}$ 與 $j$ 無關時相信是 elementary analysis 裏非常 standard 的 exercise。齊來取一些 numerical example 看看能否得到有趣的 identity。

• $c_{jn}=1/n\implies \dis \limn \prod_{j=1}^n  \bigg(1+\frac{1}{n}\bigg)=\limn \bigg(1+\frac{1}{n}\bigg)^n=e$
• If $p> 1$, $\dis c_{jn} = 1/n^p\implies \limn \prod_{j=1}^n \bigg(1+\frac{1}{n^p}\bigg) = \limn \bigg(1+\frac{1}{n^p}\bigg)^n=e^{\limn\frac{1}{n^{p-1}}}=1$
• $\dis c_{jn}= \frac{j^k}{n^{k+1}} \implies \dis \limn \prod_{j=1}^n \bigg(1+ \frac{j^k}{n^{k+1}}\bigg)=\exp \bigg(\limn \frac{\sum_{j=1}^n j^k}{n^{k+1}}\bigg)=e^{1/(k+1)}$
• $\dis c_{jn} = B_n \times \int_{j-1}^j f(x)\,dx$, then $$\limn \prod_{j=1}^n \bigg(1+B_n \int_{j-1}^j f(x)\,dx \bigg) =\exp \brac{\limn B_n\int_0^n f(x)\,dx},$$ in particular, if we take $B_n =(\int_{j-1}^j f(x)\,dx )^{-1}$, then $$\max_{1\leq j\leq n}\left| \frac{\int_{j-1}^j f(x)\,dx}{\int_0^n f(x)\,dx} \right|\to 0\implies \limn \prod_{j=1}^n \bigg(1+\frac{\int_{j-1}^j f(x)\,dx}{\int_0^n f(x)\,dx} \bigg)  = e.$$
• Many examples of infinite products converge to $1$.