Integral of a sum 3

Level pending

Let x > 1 x > 1 and m > 0 m > 0 .

If ∫ m + 1 m + 2 ∑ n = 1 ∞ n 3 x n d x = m 6 + 3 m 5 + 10 m 4 + 27 m 3 + 31 m 2 + 12 m + 2 m 3 ( m + 1 ) 3 \int_{m + 1}^{m + 2} \sum_{n = 1}^{\infty} \dfrac{n^3}{x^n} dx = \dfrac{m^6 + 3 m^5 + 10 m^4 + 27 m^3 + 31 m^2 + 12 m + 2}{m^3(m + 1)^3} , find the value of m m to seven decimal places.


The answer is 0.5819767.

Rocco Dalto by Rocco Dalto , 67, USA

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1 solution

Rocco Dalto
Jan 4, 2018

Let x > 1 x > 1 and m > 0 m > 0 .

∑ n = 1 ∞ n x n = 1 x ( 1 + 1 x + 1 x 2 + 1 x 3 + . . . ) + 1 x 2 ( 1 + 1 x + 1 x 2 + 1 x 3 + . . . ) + 1 x 3 ( 1 + 1 x + 1 x 2 + 1 x 3 + . . . ) + . . . = \sum_{n = 1}^{\infty} \dfrac{n}{x^n} = \dfrac{1}{x}(1 + \dfrac{1}{x} + \dfrac{1}{x^2} + \dfrac{1}{x^3} + ... ) + \dfrac{1}{x^2}(1 + \dfrac{1}{x} + \dfrac{1}{x^2} + \dfrac{1}{x^3} + ...) +\dfrac{1}{x^3}(1 + \dfrac{1}{x} + \dfrac{1}{x^2} + \dfrac{1}{x^3} + ...) + ... =

( x x − 1 ) ( 1 x ) ( x x − 1 ) = x ( x − 1 ) 2 . (\dfrac{x}{x - 1})(\dfrac{1}{x})(\dfrac{x}{x - 1}) = \dfrac{x}{(x - 1)^2}.

∑ n = 1 ∞ n 2 x n = ∑ j = 1 ∞ ( ∑ n = j ∞ n x n ) = \sum_{n = 1}^{\infty} \dfrac{n^2}{x^n} = \sum_{j = 1}^{\infty} (\sum_{n = j}^{\infty} \dfrac{n}{x^n}) = ∑ j = 1 ∞ ( j x j + j + 1 x j + 1 + j + 2 x j + 2 + . . . ) = \sum_{j = 1}^{\infty} (\dfrac{j}{x^j} + \dfrac{j + 1}{x^{j + 1}} + \dfrac{j + 2}{x^{j + 2}} + ... ) = ( x x − 1 ) ∑ j = 1 ∞ j x j + x ( x − 1 ) 2 ∑ j = 1 ∞ 1 x j = ( x x − 1 ) ( x ( x − 1 ) 2 ) + x ( x − 1 ) 2 ( 1 x − 1 ) = x 2 + x ( x − 1 ) 3 . (\dfrac{x}{x - 1}) \sum_{j = 1}^{\infty} \dfrac{j}{x^j} + \dfrac{x}{(x - 1)^2} \sum_{j = 1}^{\infty} \dfrac{1}{x^j} = (\dfrac{x}{x - 1}) (\dfrac{x}{(x - 1)^2}) + \dfrac{x}{(x - 1)^2} (\dfrac{1}{x - 1}) = \dfrac{x^2 + x}{(x - 1)^3}.

S ( x ) = ∑ n = 1 ∞ n 3 x n = ∑ j = 0 ∞ ( ∑ n = j ∞ n 2 x n ) = S(x) = \sum_{n = 1}^{\infty} \dfrac{n^3}{x^n} = \sum_{j = 0}^{\infty} (\sum_{n = j}^{\infty} \dfrac{n^2}{x^{n}}) =

∑ j = 0 ∞ ( j 2 x j ( ∑ k = 0 ∞ 1 x k ) + 2 j x j \sum_{j = 0}^{\infty} (\dfrac{j^2}{x^j} (\sum_{k = 0}^{\infty} \dfrac{1}{x^k}) + \dfrac{2j}{x^j} ( ∑ k = 1 ∞ k x k ) (\sum_{k = 1}^{\infty} \dfrac{k}{x^k}) + 1 x j ( ∑ k = 1 ∞ k 2 x k ) ) = \dfrac{1}{x^j} (\sum_{k = 1}^{\infty} \dfrac{k^2}{x^k}) ) =

x x − 1 ∑ j = 1 ∞ j 2 x j + 2 x ( x − 1 ) 2 ∑ j = 1 ∞ j x j + x 2 + x ( x − 1 ) 3 ∑ j = 1 ∞ 1 x j = \dfrac{x}{x - 1} \sum_{j = 1}^{\infty} \dfrac{j^2}{x^j} + \dfrac{2x}{(x - 1)^2} \sum_{j = 1}^{\infty} \dfrac{j}{x^j} + \dfrac{x^2 + x}{(x - 1)^3} \sum_{j = 1}^{\infty} \dfrac{1}{x^j} =

( x x − 1 ) ( x 2 + x ( x − 1 ) 3 ) + ( 2 x ( x − 1 ) 2 ) ( x ( x − 1 ) 2 ) + ( x 2 + x ( x − 1 ) 3 ) ( 1 x ) = (\dfrac{x}{x - 1}) (\dfrac{x^2 + x}{(x - 1)^3}) + (\dfrac{2x}{(x - 1)^2}) (\dfrac{x}{(x - 1)^2}) + (\dfrac{x^2 + x}{(x - 1)^3}) (\dfrac{1}{x}) = x 3 + 4 x 2 + x ( x − 1 ) 4 \dfrac{x^3 + 4x^2 + x}{(x - 1)^4} .

Let u = x − 1 ⟹ d u = d x ⟹ ∫ m + 1 m + 2 S ( x ) d x = ∫ m m + 1 u 3 + 7 u 2 + 12 u + 6 u 4 d u = ∫ m m + 1 ( 1 u + 7 u − 2 + 12 u − 3 + 6 u − 4 ) d u u = x - 1 \implies du = dx \implies \int_{m + 1}^{m + 2} S(x) dx = \int_{m}^{m + 1} \dfrac{u^3 + 7u^2 + 12u + 6}{u^4} du = \int_{m}^{m + 1} (\dfrac{1}{u} + 7u^{-2} + 12u^{-3} + 6u^{-4}) du

= ( ln ⁡ ( u ) − 7 u − 6 u 2 − 2 u 3 ) ∣ m m + 1 = = (\ln(u) - \dfrac{7}{u} - \dfrac{6}{u^2} - \dfrac{2}{u^3})|_{m}^{m + 1} = ln ⁡ ( m + 1 m ) + 7 m 4 + 26 m 3 + 31 m 2 + 12 m + 2 m 3 ( m + 1 ) 3 = \ln(\dfrac{m + 1}{m}) + \dfrac{7m^4 + 26m^3 + 31m^2 + 12m + 2}{m^3 (m + 1)^3} = m 6 + 3 m 5 + 10 m 4 + 27 m 3 + 31 m 2 + 12 m + 2 m 3 ( m + 1 ) 3 = \dfrac{m^6 + 3 m^5 + 10 m^4 + 27 m^3 + 31 m^2 + 12 m + 2}{m^3(m + 1)^3} = m 6 + 3 m 5 + 3 m 4 + m 3 + 7 m 4 + 26 m 3 + 31 m 2 + 12 m + 2 m 3 ( m + 1 ) 3 = \dfrac{m^6 + 3m^5 + 3m^4 + m^3 + 7m^4 + 26m^3 + 31m^2 + 12m + 2}{m^3 (m + 1)^3} =

m 3 ( m + 1 ) 3 + 7 m 4 + 26 m 3 + 31 m 2 + 12 m + 2 m 3 ( m + 1 ) 3 = \dfrac{m^3 (m + 1)^3 + 7m^4 + 26m^3 + 31m^2 + 12m + 2}{m^3 (m + 1)^3} = 1 + 7 m 4 + 26 m 3 + 31 m 2 + 12 m + 2 m 3 ( m + 1 ) 3 ⟹ ln ⁡ ( m + 1 m ) = 1 ⟹ m + 1 m = e ⟹ m = 1 e − 1 = 0.5819767 1 + \dfrac{7m^4 + 26m^3 + 31m^2 + 12m + 2}{m^3 (m + 1)^3} \implies \ln(\dfrac{m + 1}{m}) = 1 \implies \dfrac{m + 1}{m} = e \implies m = \dfrac{1}{e - 1} = \boxed{0.5819767} to seven decimal places.

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