ข้อสอบ PAT 1 - มีนาคม 2556

$$\begin{gathered} \mathrm{จากโจทย์} {\mathrm{a}}_{\mathrm{n}}=\frac{1}{4+8+12+...+4\mathrm{n}} ; \mathrm{n}=1,2,3,... \\ {\mathrm{a}}_{\mathrm{n}}=\frac{1}{4\left(1+2+3+...+\mathrm{n}\right)} \\ =\frac{1}{4\left[{\displaystyle \frac{\mathrm{n}}{2}\left(\mathrm{n}+1\right)}\right]}=\frac{1}{2\mathrm{n}\left(\mathrm{n}+1\right)} \\ \mathrm{ผลบวกของอนุกรม} {\mathrm{a}}_1+{\mathrm{a}}_2+{\mathrm{a}}_3+...\mathrm{คือ} {\mathrm{S}}_{\mathrm{n}} \end{gathered}$$
$$\begin{gathered} \mathrm{จากสูตร} {\mathrm{S}}_{\mathrm{n}}=\sum _{\mathrm{n}=1}^{\infty }{\mathrm{a}}_{\mathrm{n}} \\ {\mathrm{S}}_{\infty }=\sum _{\mathrm{n}=1}^{\infty }\left(\frac{1}{2\mathrm{n}\left(\mathrm{n}+1\right)}\right) \\ =\frac{1}{2}\sum _{\mathrm{n}=1}^{\infty }\left(\frac{1}{\mathrm{n}\left(\mathrm{n}+1\right)}\right) \\ =\frac{1}{2}\sum _{\mathrm{n}=1}^{\infty }\left(\frac{1}{\mathrm{n}}-\frac{1}{\mathrm{n}+1}\right) \end{gathered}$$
$=\frac{1}{2}\left[\left(\frac{1}{1}-\cancel{\frac{1}{2}}\right)+\left(\cancel{\frac{1}{2}}-\cancel{\frac{1}{3}}\right)+\left(\cancel{\frac{1}{3}}-\cancel{\frac{1}{4}}\right)+...+\left(\cancel{\frac{1}{\mathrm{n}}}-\frac{1}{\mathrm{n}+1}\right)\right]$

$$\begin{gathered} {\mathrm{S}}_{\infty }=\frac{1}{2}\left[1-\frac{1}{\mathrm{n}+1}\right] \\ \mathrm{จากสูตร} {\mathrm{S}}_{\infty }=\underset{\mathrm{n}\to \infty }{\lim }{\mathrm{S}}_{\mathrm{n}} \\ =\underset{\mathrm{n}\to \infty }{\lim }\frac{1}{2}\left[1-\frac{1}{\mathrm{n}+1}\right] \end{gathered}$$

$$\begin{gathered} =\frac{1}{2}\underset{\mathrm{n}\to \infty }{\lim }\left[1-\cancel{\frac{1}{\mathrm{n}+1}}\right] \\ =\frac{1}{2}\left[1-0\right] \\ {\mathrm{S}}_{\infty }=\frac{1}{2} \\ \mathrm{ดังนั้น} \mathrm{ผลบวกของอนุกรม} {\mathrm{a}}_1+{\mathrm{a}}_2+{\mathrm{a}}_3+...\mathrm{เท่ากับ} \frac{1}{2} \end{gathered}$$