Integral of a sum 2

Calculus Level 4

If ∫ m + 1 m + 2 ∑ n = 1 ∞ n 2 x n d x = m 4 + 2 m 3 + 4 m 2 + 5 m + 1 m 2 ( m + 1 ) 2 \displaystyle \int_{m + 1}^{m + 2} \sum_{n = 1}^{\infty} \dfrac{n^2}{x^n} dx = \dfrac{m^4 + 2 m^3 + 4m^2 + 5 m + 1}{m^2(m + 1)^2} , for x > 1 x > 1 and m > 0 m > 0 , 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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2 solutions

Consider the following series for ∣ u ∣ < 1 |u| < 1 .

∑ n = 0 ∞ u n = 1 1 − u Differentiate both sides w.r.t. u . ∑ n = 1 ∞ n u n − 1 = 1 ( 1 − u ) 2 Multiplied both sides by u . ∑ n = 1 ∞ n u n = u ( 1 − u ) 2 Differentiate both sides w.r.t. u . ∑ n = 1 ∞ n 2 u n − 1 = 1 + u ( 1 − u ) 3 Multiplied both sides by u . ∑ n = 1 ∞ n 2 u n = u ( 1 + u ) ( 1 − u ) 3 Replace u with 1 x ∑ n = 1 ∞ n 2 x n = x 2 + x ( x − 1 ) 3 \begin{aligned} \sum_{\color{#3D99F6}n=0}^\infty u^n & = \frac 1{1-u} & \small \color{#3D99F6} \text{Differentiate both sides w.r.t. }u. \\ \sum_{\color{#D61F06}n=1}^\infty nu^{n-1} & = \frac 1{(1-u)^2} & \small \color{#3D99F6} \text{Multiplied both sides by }u. \\ \sum_{n=1}^\infty nu^n & = \frac u{(1-u)^2} & \small \color{#3D99F6} \text{Differentiate both sides w.r.t. }u. \\ \sum_{n=1}^\infty n^2u^{n-1} & = \frac {1+u}{(1-u)^3} & \small \color{#3D99F6} \text{Multiplied both sides by }u. \\ \sum_{n=1}^\infty n^2u^n & = \frac {u(1+u)}{(1-u)^3} & \small \color{#3D99F6} \text{Replace }u \text{ with }\frac 1x \\ \sum_{n=1}^\infty \frac {n^2}{x^n} & = \frac {x^2+x}{(x-1)^3} \end{aligned}

Therefore, we have:

I = ∫ m + 1 m + 2 ∑ n = 1 ∞ n 2 x n d x = ∫ m + 1 m + 2 x 2 + x ( x − 1 ) 3 d x = ∫ m + 1 m + 2 x 2 − 2 x + 1 + 3 x − 3 + 2 ( x − 1 ) 3 d x = ∫ m + 1 m + 2 ( 1 x − 1 + 3 ( x − 1 ) 2 + 2 ( x − 1 ) 3 ) d x = ln ⁡ ( x − 1 ) − 3 x − 1 − 1 ( x − 1 ) 2 ∣ m + 1 m + 2 = ln ⁡ ( x − 1 ) − 3 x − 2 ( x − 1 ) 2 ∣ m + 1 m + 2 = ln ⁡ ( m + 1 m ) − 3 m + 4 ( m + 1 ) 2 + 3 m + 1 m 2 = ln ⁡ ( m + 1 m ) + 3 m 2 + 5 m + 1 m 2 ( m + 1 ) 2 = ln ⁡ ( m + 1 m ) + m 2 + 2 m 3 + 4 m 2 + 5 m + 1 m 2 ( m + 1 ) 2 − m 2 + 2 m 3 + m 2 m 2 ( m + 1 ) 2 = ln ⁡ ( m + 1 m ) + m 2 + 2 m 3 + 4 m 2 + 5 m + 1 m 2 ( m + 1 ) 2 − 1 \begin{aligned} I & = \int_{m+1}^{m+2} \sum_{n=1}^\infty \frac {n^2}{x^n} dx \\ & = \int_{m+1}^{m+2} \frac {x^2+x}{(x-1)^3} dx \\ & = \int_{m+1}^{m+2} \frac {x^2-2x+1+3x-3+2}{(x-1)^3} dx \\ & = \int_{m+1}^{m+2} \left(\frac 1{x-1} + \frac 3{(x-1)^2} + \frac 2{(x-1)^3} \right) dx \\ & = \ln (x-1) - \frac 3{x-1} - \frac 1{(x-1)^2} \ \bigg|_{m+1}^{m+2} \\ & = \ln (x-1) - \frac {3x-2}{(x-1)^2} \ \bigg|_{m+1}^{m+2} \\ & = \ln \left(\frac {m+1}m \right) - \frac {3m+4}{(m+1)^2} + \frac {3m+1}{m^2} \\ & = \ln \left(\frac {m+1}m \right) + \frac {3m^2+5m+1}{m^2(m+1)^2} \\ & = \ln \left(\frac {m+1}m \right) + \frac {{\color{#3D99F6}m^2+2m^3+4m^2}+5m+1}{m^2(m+1)^2} - \frac {\color{#D61F06}m^2+2m^3+m^2}{m^2(m+1)^2} \\ & = {\color{#D61F06}\ln \left(\frac {m+1}m \right)} + \frac {m^2+2m^3+4m^2+5m+1}{m^2(m+1)^2} \color{#D61F06} - 1 \end{aligned}

This implies that:

ln ⁡ ( m + 1 m ) = 1 m + 1 m = e ⟹ m = 1 e − 1 ≈ 0.5819767 \begin{aligned} \ln \left(\frac {m+1}m \right) & = 1 \\\frac {m+1}m & = e \\ \implies m & = \frac 1{e-1} \approx \boxed{0.5819767} \end{aligned}

2 Helpful 0 Interesting 1 Brilliant 0 Confused

I hope you know there's already a proof on this special sums on a generalised aspect ∑ n = 1 ∞ n x k n = k A n ( k ) ( k − 1 ) x + 1 \displaystyle\sum_{n=1}^\infty\dfrac{n^x}{k^n}=\dfrac{kA_n(k)}{(k-1)^{x+1}} where A n ( k ) A_n(k) is the n n th Eulerian polynomial.

Syed Shahabudeen - 3 years, 4 months ago
Rocco Dalto
Jan 3, 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}

S ( x ) = ∑ n = 1 ∞ n 2 x n = ∑ j = 1 ∞ ( ∑ n = j ∞ n x n ) = S(x) = \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}

For ∫ m + 1 m + 2 S ( x ) d x = ∫ m + 1 m + 2 x 2 + x ( x − 1 ) 3 \int_{m + 1}^{m + 2} S(x) dx = \int_{m + 1}^{m + 2} \dfrac{x^2 + x}{(x - 1)^3}

Let u = x − 1 ⟹ d u = d x ⟹ ∫ m + 1 m + 2 S ( x ) d x = ∫ m m + 1 ( 1 u + 3 u 2 + 2 u 3 ) d u = u = x - 1 \implies du = dx \implies \int_{m + 1}^{m + 2} S(x) dx = \int_{m}^{m + 1} (\dfrac{1}{u} + \dfrac{3}{u^2} + \dfrac{2}{u^3}) du = ( ln ⁡ ( u ) − 3 u − 1 u 2 ) ∣ m m + 1 = ln ⁡ ( m + 1 m ) + 3 m 2 + 5 m + 1 ( m + 1 ) 2 m 2 = m 4 + 2 m 3 + 4 m 2 + 5 m + 1 m 2 ( m + 1 ) 2 = (\ln(u) - \dfrac{3}{u} - \dfrac{1}{u^2})|_{m}^{m + 1} = \ln(\dfrac{m + 1}{m}) + \dfrac{3 m^2 + 5 m + 1}{(m + 1)^2 m^2} = \dfrac{m^4 + 2 m^3 + 4m^2 + 5 m + 1}{m^2(m + 1)^2} =

m 4 + 2 m 3 + m 2 + 3 m 2 + 5 m + 1 ( m + 1 ) 2 m 2 = ( m + 1 ) 2 m 2 + 3 m 2 + 5 m + 1 ( m + 1 ) 2 m 2 = 1 + 3 m 2 + 5 m + 1 ( m + 1 ) 2 m 2 \dfrac{m^4 + 2 m^3 + m^2 + 3 m^2 + 5 m + 1}{(m + 1)^2 m^2} = \dfrac{(m + 1)^2 m^2 + 3 m^2 + 5 m + 1}{(m + 1)^2 m^2} = 1 + \dfrac{3 m^2 + 5 m + 1}{(m + 1)^2 m^2}

⟹ ln ⁡ ( m + 1 m ) = 1 ⟹ m + 1 m = e ⟹ m = 1 e − 1 = 0.5819767 \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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