Cho $S = \dfrac{1}{3} – \dfrac{2}{3^2} + \dfrac{3}{3^3} – \dfrac{4}{3^4} +…+ \dfrac{99}{3^{99}} – \dfrac{100}{3^{100}}$
So sánh $S$ với $\dfrac{1}{5}$
Cho $S = \dfrac{1}{3} – \dfrac{2}{3^2} + \dfrac{3}{3^3} – \dfrac{4}{3^4} +…+ \dfrac{99}{3^{99}} – \dfrac{100}{3^{100}}$
So sánh $S$ với $\dfrac{1}{5}$
`S = 1/3 – 2/3^2 + 3/3^3 – … – 100/3^100`
`⇒ 3S = 1 – 2/3 + 3/3^2 – …. – 100/3^99`
`⇒ 3S + S = 1 – 1/3 + 1/3^2 – … – 1/3^99 – 100/3^100`
Đặt `D = 1 – 1/3 + 1/3^2 – … – 1/3^99`
`⇒ 3D = 3 – 1 + 1/3 – 1/3^2 + … – 1/3^98`
`⇒ 4D = 3 – 1/3^99`
`⇒ D = (3 – 1/3^99). 1/4`
Thay `D` vào `S` ta có:
`4C = 1/4. (3 – 1/3^99) – 100/3^100`
`⇒ 4S = 3/4 – 1/(4. 3^99) – 100/3^100`
`⇒ S = 1/4. (3/4 – 1/(4. 3^99) – 100/3^100)`
`⇒ S = 3/16 – 1/(4^2. 3^99) – 25/3^100`
`⇒ S = 3/16 – (1/(4^2. 3^99) + 25/3^100)`
`⇒ S < 3/16` mà `3/16 < 1/5`
`⇒ S < 1/5`