Sequences of functions

Level pending

If 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)} ,

find ∫ n − 1 ∞ S n ( x ) d x − ∫ 2 n + 1 ∞ T n ( x ) d x \int_{n - 1}^{\infty} S_{n}(x) dx - \int_{2n + 1}^{\infty} T_{n}(x) dx .


The answer is 0.

Rocco Dalto by Rocco Dalto , 67, USA

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Rocco Dalto
Jul 18, 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}

⟹ ∫ n − 1 ∞ S n ( x ) d x = ln ⁡ ( 1 − n x + n + 1 ) ∣ n − 1 ∞ = − ln ⁡ ( 1 2 ) = ln ⁡ ( 2 ) \implies \int_{n - 1}^{\infty} S_{n}(x) dx = \ln(1 - \dfrac{n}{x + n + 1})|_{n - 1}^{\infty} = -\ln(\dfrac{1}{2}) = \ln(2) and ∫ 2 n + 1 ∞ T n ( x ) d x = ln ⁡ ( 1 − n x − 1 ) ∣ 2 n + 1 ∞ = − ln ⁡ ( 1 2 ) = ln ⁡ ( 2 ) \int_{2n + 1}^{\infty} T_{n}(x) dx = \ln(1 - \dfrac{n}{x - 1})|_{2n + 1}^{\infty} = -\ln(\dfrac{1}{2}) = \ln(2)

⟹ ∫ n − 1 ∞ S n ( x ) d x − ∫ 2 n + 1 ∞ T n ( x ) d x = 0 \implies \int_{n - 1}^{\infty} S_{n}(x) dx - \int_{2n + 1}^{\infty} T{n}(x) dx = 0 .

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...