Infinity sum

Calculus Level 5

1 1 ⋅ 2 ⋅ 3 ⋅ 4 ⋅ 5 + 1 3 ⋅ 4 ⋅ 5 ⋅ 6 ⋅ 7 + 1 5 ⋅ 6 ⋅ 7 ⋅ 8 ⋅ 9 + ⋯ = ? \frac{ 1 } { 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5} +\frac{ 1} {3 \cdot 4\cdot 5\cdot 6\cdot 7} + \frac{ 1}{ 5 \cdot 6 \cdot 7 \cdot 8 \cdot9 } + \cdots = \, ?

Give your answer to 3 significant figures.


The answer is 0.00882.

A Former Brilliant Member by A Former Brilliant Member

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2 solutions

First Last
Oct 15, 2016

Using partial fractions, for the sum S = ∑ n = 1 ∞ 1 ( 2 n − 1 ) ( 2 n ) ( 2 n + 1 ) ( 2 n + 2 ) ( 2 n + 3 ) = \displaystyle\textbf{S} = \sum_{n=1}^{\infty}\frac{1}{(2n-1)(2n)(2n+1)(2n+2)(2n+3)} =

1 24 ∑ n = 1 ∞ ( − 2 n + − 2 n + 1 + 1 2 n − 1 + 6 2 n + 1 + 1 2 n + 3 ) \displaystyle\frac{1}{24}\sum_{n=1}^{\infty}( \frac{-2}{n} + \frac{-2}{n+1}+\frac{1}{2n-1}+\frac{6}{2n+1}+\frac{1}{2n+3}) , disregard the 1 24 \frac{1}{24} until the end.

∑ n = 1 ∞ − 2 n + − 2 n + 1 = − 4 ( 1 + 1 2 + 1 3 + . . . ) + 2 \displaystyle\sum_{n=1}^{\infty}\frac{-2}{n} + \frac{-2}{n+1} = -4(1+\frac{1}{2}+\frac{1}{3}+...)+2

∑ n = 1 ∞ 1 2 n − 1 + 6 2 n + 1 + 1 2 n + 3 = 1 + 1 3 + 6 3 + 8 ( 1 5 + 1 7 + . . . ) \displaystyle\sum_{n=1}^{\infty}\frac{1}{2n-1}+\frac{6}{2n+1}+\frac{1}{2n+3} = 1 + \frac{1}{3}+\frac{6}{3}+8(\frac{1}{5}+\frac{1}{7}+...)

16 3 + 8 ( 1 5 + 1 7 + . . . ) − 4 ( 1 + 1 2 + 1 3 + . . . ) = \displaystyle \frac{16}{3} + 8(\frac{1}{5}+\frac{1}{7}+...) -4(1+\frac{1}{2}+\frac{1}{3}+...) =

16 3 − 8 ( 1 + 1 3 ) + 8 ( 1 + 1 3 + 1 5 + . . . ) − 8 [ 1 2 ( 1 + 1 2 + 1 3 + . . . ) ] = \displaystyle\frac{16}{3} - 8(1 + \frac{1}{3}) + 8(1+\frac{1}{3} + \frac{1}{5} + ...) - 8[\frac{1}{2}(1 + \frac{1}{2} + \frac{1}{3} + ...)]=

− 16 3 + 8 ( 1 − 1 2 + 1 3 − . . . ) = − 16 3 + 8 ln ⁡ 2 \displaystyle\frac{-16}{3} + 8(1 -\frac{1}{2}+\frac{1}{3}-...) = \frac{-16}{3}+8\ln{2}

Bring back the 1 24 , S = ln ⁡ 2 3 − 2 9 ≈ . 008826838 \displaystyle\frac{1}{24}, \quad\textbf{S} = \frac{\ln{2}}{3}-\frac{2}{9}\approx.008826838

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An interesting way to solve series like this would be to set up the sum as ∑ n = 1 ∞ ∫ 0 1 ∫ 0 a ∫ 0 b . . . y 2 n − 2 d y d d d c d b d a = ∫ 0 1 ∫ 0 a ∫ 0 b . . . ∑ n = 1 ∞ y 2 n − 2 d y d d d c d b d a \displaystyle\sum_{n=1}^{\infty}\int_{0}^{1}\int_{0}^{a}\int_{0}^{b}...y^{2n-2}dydddcdbda = \int_{0}^{1}\int_{0}^{a}\int_{0}^{b}...\sum_{n=1}^{\infty}y^{2n-2}\,dydddcdbda . The sum is just a geometric progression.

Much more applicable for smaller cases.

First Last - 4 years, 8 months ago

This is a good start. However, your explanation has a slight error.

  1. For non- absolutely convergent sequences, we cannot just break up the terms and wait for them to cancel out. Care has to be taken when justifying the cancellation.

Keep writing more solutions and you will get the hang of this!

Calvin Lin Staff - 4 years, 8 months ago
Hassan Abdulla
Jul 18, 2019

S = ∑ k = 0 ∞ 1 ( 2 k + 1 ) ( 2 k + 2 ) ( 2 k + 3 ) ( 2 k + 4 ) ( 2 k + 5 ) S = 1 24 ∑ k = 0 ∞ 1 2 k + 1 − 4 2 k + 2 + 6 2 k + 3 − 4 2 k + 4 + 1 2 k + 5 Using partial fractions S = 1 24 ∑ k = 0 ∞ ( ∫ 0 1 ( x 2 k − 4 x 2 k + 1 + 6 x 2 k + 2 − 4 x 2 k + 3 + x 2 k + 4 ) d x ) manipulate the fractions S = 1 24 ∑ k = 0 ∞ ( ∫ 0 1 x 2 k ⋅ ( 1 − 4 x + 6 x 2 − 4 x 3 + x 4 ) d x ) take x 2 k as common factor S = 1 24 ∑ k = 0 ∞ ( ∫ 0 1 x 2 k ⋅ ( 1 − x ) 4 d x ) S = 1 24 ∫ 0 1 ( ( 1 − x ) 4 ⋅ ∑ k = 0 ∞ x 2 k ) d x interchange integral with sum S = 1 24 ∫ 0 1 ( 1 − x ) 4 1 − x 2 d x = 1 24 ∫ 0 1 ( 1 − x ) 3 1 + x d x ∑ k = 0 ∞ x 2 k is geometric series since ∣ x ∣ < 1 S = 1 24 ∫ 0 1 ( − x 2 + 4 x − 7 + 8 x + 1 ) d x dividing the polynomials S = 1 24 ( − x 3 3 + 2 x 2 − 7 x + 8 ln ⁡ ( x + 1 ) ∣ 0 1 ) S = 1 24 ( 8 ln ⁡ ( 2 ) − 16 3 ) = ln ⁡ ( 2 ) 3 − 2 9 ≈ 0.008826 \begin{aligned} &S=\sum_{k=0}^\infty \frac{1}{(2k+1)(2k+2)(2k+3)(2k+4)(2k+5)}\\ &S=\frac{1}{24}\sum_{k=0}^\infty \frac{1}{2k+1} - \frac{4}{2k+2} + \frac{6}{2k+3} - \frac{4}{2k+4} + \frac{1}{2k+5} && {\color{#D61F06} \text{Using partial fractions}}\\ &S=\frac{1}{24}\sum_{k=0}^\infty \left ( \int_0^1 \left ( x^{2k}-4x^{2k+1}+6x^{2k+2}-4x^{2k+3}+x^{2k+4} \right )\,dx \right ) && {\color{#D61F06} \text{manipulate the fractions}} \\ &S=\frac{1}{24}\sum_{k=0}^\infty \left ( \int_0^1 x^{2k} \cdot \left ( 1-4x+6x^2-4x^3+x^4 \right )\,dx \right ) && {\color{#D61F06} \text{take } x^{2k} \text{ as common factor}} \\ &S=\frac{1}{24}\sum_{k=0}^\infty \left ( \int_0^1 x^{2k} \cdot \left ( 1-x \right )^4\,dx \right ) \\ &S=\frac{1}{24}\int_0^1 \left ( \left ( 1-x \right )^4 \cdot \sum_{k=0}^\infty x^{2k} \right )\,dx && {\color{#D61F06} \text{interchange integral with sum}}\\ &S=\frac{1}{24}\int_0^1 \frac{\left ( 1-x \right )^4}{1-x^2}\,dx=\frac{1}{24}\int_0^1 \frac{\left ( 1-x \right )^3}{1+x}\,dx && {\color{#D61F06} \sum _{k=0}^\infty x^{2k}\text{ is geometric series since} \left | x \right | < 1} \\ &S=\frac{1}{24}\int_0^1 \left ( -x^2+4x-7+\frac{8}{x+1} \right )\,dx && {\color{#D61F06} \text{ dividing the polynomials}} \\ &S=\frac{1}{24} \left ( \left . \frac{-x^3}{3} + 2x^2 - 7x + 8 \ln(x+1) \right |_0^1 \right )\\ &S=\frac{1}{24} \left ( 8\ln(2)-\frac{16}{3} \right )=\frac{\ln(2)}{3}-\frac{2}{9}\approx 0.008826 \end{aligned}

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