[Euler-Fermat]

ให้ $a_n = 2+4+6+.....+2n$

$b_n = a_1+a_2+a_3...+a_n$

หา $\lim_{n \to \infty}[\frac{2}{b_1}+\frac{3}{b_2}+ \frac{4}{b_3}+...+\frac{n+1}{b_n}]$

ได้ว่า $a_n = n(n+1)$

$b_n = \sum_{k = 1}^{n} k(k+1) = \dfrac{n(n+1)(n+2)}{3}$

$\lim_{n \to \infty}[\frac{2}{b_1}+\frac{3}{b_2}+ \frac{4}{b_3}+...+\frac{n+1}{b_n}]$

$= \lim_{n \to \infty}[\sum_{k = 1}^{n} \dfrac{k+1}{b_k}] $

ให้ $c_n = \dfrac{n+1}{b_n} = \dfrac{3}{n(n+2)} = \dfrac{3}{2}[\dfrac{1}{n}-\dfrac{1}

{n+2}]$

$\therefore \lim_{n \to \infty}[\sum_{k = 1}^{n} \dfrac{k+1}{b_k}] $

$= \lim_{n \to \infty}[\sum_{k = 1}^{n} c_k]$

$= \lim_{n \to \infty}[\sum_{k = 1}^{n}\dfrac{3}{2}[\dfrac{1}{k}-\dfrac{1}{k+2}] ] $

$= \lim_{n \to \infty}[\dfrac{3}{2}[\dfrac{3}{2}-\dfrac{1}{n+1}-\dfrac{1}{n+2}]]$

$= \dfrac{9}{4}$