Sequences of functions 2

Calculus Level 4

Let S n ( x ) = ∑ j = 1 n 1 ( x + j ) ( x + j + 1 ) S_{n}(x) = \sum_{j = 1}^{n} \dfrac{1}{(x + j)(x + j + 1)} and T n ( x ) = ∑ j = 1 n 1 ( x − j ) ( x − j − 1 ) T_{n}(x) = \sum_{j = 1}^{n} \dfrac{1}{(x - j)(x - j - 1)} .

When revolved about the x x axis let V 1 V_{1} be the volume of the region bounded by S n ( x ) S_{n}(x) and the x x axis on [ n − 1 , ∞ ) [n - 1,\infty) and V 2 V_{2} be the volume of the region bounded by T n ( x ) T_{n}(x) and the x x axis on [ 2 n + 1 , ∞ ) [2n + 1,\infty) .

Find V 1 − V 2 V_{1} - V_{2} .


The answer is 0.

Rocco Dalto by Rocco Dalto , 67, USA

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1 solution

Rocco Dalto
Jul 19, 2018

S n ( x ) = ∑ j = 1 n 1 ( x + j ) ( x + j + 1 ) = ∑ j = 1 n 1 x + j − 1 x + j + 1 = 1 x + 1 − 1 x + n + 1 S_{n}(x) = \sum_{j = 1}^{n} \dfrac{1}{(x + j)(x + j + 1)} = \sum_{j = 1}^{n} \dfrac{1}{x + j} - \dfrac{1}{x + j + 1} = \dfrac{1}{x + 1} - \dfrac{1}{x + n + 1}

and

T n ( x ) = ∑ j = 1 n 1 ( x − j ) ( x − j − 1 ) = ∑ j = 1 n 1 x − j − 1 − 1 x − j = 1 x − n − 1 − 1 x − 1 T_{n}(x) = \sum_{j = 1}^{n} \dfrac{1}{(x - j)(x - j - 1)} = \sum_{j = 1}^{n} \dfrac{1}{x - j - 1} - \dfrac{1}{x - j} = \dfrac{1}{x - n - 1} - \dfrac{1}{x - 1}

V 1 = π ∫ n − 1 ∞ S n ( x ) d x = π ∫ n − 1 ∞ ( 1 x + 1 − 1 x + n + 1 ) 2 d x = V_{1} = \pi\int_{n - 1}^{\infty} S_{n}(x) dx = \pi\int_{n - 1}^{\infty} (\dfrac{1}{x + 1} - \dfrac{1}{x + n + 1})^2 dx = π ∫ n − 1 ∞ ( ( x + 1 ) − 2 − 2 n ( 1 x + 1 − 1 x + n + 1 ) + ( x + n + 1 ) − 2 ) d x = \pi\int_{n - 1}^{\infty} ((x + 1)^{-2} - \dfrac{2}{n}(\dfrac{1}{x + 1} - \dfrac{1}{x + n + 1}) + (x + n + 1)^{-2}) dx = π ( − 1 x + 1 − 2 n ( ln ⁡ ( 1 − n x + n + 1 ) ) − 1 x + n + 1 ) ∣ n − 1 ∞ = \pi(-\dfrac{1}{x + 1} - \dfrac{2}{n}(\ln(1 - \dfrac{n}{x + n + 1})) - \dfrac{1}{x + n + 1})|_{n - 1}^{\infty} = π n ( 3 2 + 2 ln ⁡ ( 1 2 ) ) \boxed{\dfrac{\pi}{n}(\dfrac{3}{2} + 2\ln(\dfrac{1}{2}))}

V 2 = π ∫ 2 n + 1 ∞ T n ( x ) d x = π ∫ 2 n + 1 ∞ ( 1 x − n − 1 − 1 x − n + 1 ) 2 d x = V_{2} = \pi\int_{2n + 1}^{\infty} T_{n}(x) dx = \pi\int_{2n + 1}^{\infty} (\dfrac{1}{x - n - 1} - \dfrac{1}{x - n + 1})^2 dx = π ∫ 2 n + 1 ∞ ( ( x − n − 1 ) − 2 − 2 n ( 1 x − n − 1 − 1 x − 1 ) + ( x − 1 ) − 2 ) d x = \pi\int_{2n + 1}^{\infty} ((x - n - 1)^{-2} - \dfrac{2}{n}(\dfrac{1}{x - n - 1} - \dfrac{1}{x - 1}) + (x - 1)^{-2}) dx = π ( − 1 x − n − 1 − 2 n ( ln ⁡ ( 1 − n x − 1 ) ) − 1 x − 1 ) ∣ 2 n + 1 ∞ = \pi(-\dfrac{1}{x - n - 1} - \dfrac{2}{n}(\ln(1 - \dfrac{n}{x - 1})) - \dfrac{1}{x - 1})|_{2n + 1}^{\infty} = π n ( 3 2 + 2 ln ⁡ ( 1 2 ) ) \boxed{\dfrac{\pi}{n}(\dfrac{3}{2} + 2\ln(\dfrac{1}{2}))}

⟹ V 1 − V 2 = 0 \implies \boxed{V_{1} - V_{2} = 0}

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