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Calculus Level 5

∑ n = 0 ∞ 1 ( 4 n + 1 ) ( 4 n + 3 ) ( n + 1 ) \large \sum _{ n=0 }^{ \infty }{ \dfrac { 1 }{ ( 4n+1 ) ( 4n+3 ) ( n+1 ) } }

If the series above can be expressed in the form

π A B − ln ⁡ C , \dfrac {\pi ^A}B - \ln C,

where A , B A,B and C C are positive integers, find A + B + C A+B+C .


The answer is 6.

Aditya Kumar by Aditya Kumar , 22, India

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1 solution

Pi Han Goh
Feb 19, 2016

Synopsis : We convert the series it into a simple integral of a sum of a convergent infinite geometric progression, we then evaluate this integral by partial fractions decomposition .

∑ n = 0 ∞ 1 ( 4 n + 1 ) ( 4 n + 3 ) ( n + 1 ) = 2 3 ∑ n = 0 ∞ [ 1 4 n + 1 − 3 4 n + 3 + 2 4 n + 4 ] , Partial fractions - cover up rule = 2 3 ∑ n = 0 ∞ ∫ 0 1 ( x 4 n − 3 x 4 n + 2 + 2 x 4 n + 3 ) d x = 2 3 ∫ 0 1 1 − 3 x 2 + 2 x 3 1 − x 4 d x , Geometric series = 2 3 ∫ 0 1 ( 1 − x ) 2 ( 2 x + 1 ) ( 1 − x ) ( 1 + x ) ( x 2 + 1 ) d x = − 2 3 ∫ 0 1 ( x − 1 ) ( 2 x + 1 ) ( 1 + x ) ( x 2 + 1 ) d x Apply partial fractions = − 2 3 ∫ 0 1 ( x − 2 x 2 + 1 + 1 x + 1 ) d x = − 2 3 ∫ 0 1 ( x x 2 + 1 − 2 x 2 + 1 + 1 x + 1 ) d x = − 2 3 [ 1 2 ln ⁡ ∣ x 2 + 1 ∣ − 2 tan ⁡ − 1 ( x ) + ln ⁡ ∣ x + 1 ∣ ] 0 1 = − 2 3 ( 3 2 ln ⁡ 2 − π 2 ) = π 3 − ln ⁡ 2 \begin{array} {r l l } \displaystyle & \displaystyle \sum_{n=0}^\infty \dfrac1{(4n+1)(4n+3)(n+1)} \\ \displaystyle &= \displaystyle \dfrac23 \sum_{n=0}^\infty \left [\dfrac1{4n+1} - \dfrac3{4n+3} + \dfrac2{4n+4} \right ] , & \quad \text{ Partial fractions - cover up rule} \\ \displaystyle &= \displaystyle \dfrac23 \sum_{n=0}^\infty \int_0^1 \left( x^{4n} - 3x^{4n+2} + 2x^{4n+3} \right) \, dx \\ \displaystyle &= \displaystyle \dfrac23 \int_0^1 \dfrac{1-3x^2 + 2x^3}{1-x^4} \, dx , & \quad \text{ Geometric series} \\ \displaystyle &= \displaystyle \dfrac23 \int_0^1 \dfrac{(1-x)^{\cancel2}(2x+1)}{\cancel{(1-x)}(1+x)(x^2+1)} \, dx \\ \displaystyle &= \displaystyle -\dfrac23 \int_0^1 \dfrac{(x-1)(2x+1)}{(1+x)(x^2+1)} \, dx & \quad \text{ Apply partial fractions} \\ \displaystyle &= \displaystyle -\dfrac23 \int_0^1 \left( \dfrac{x-2}{x^2+1} + \dfrac1{x+1} \right) \, dx \\ \displaystyle &= \displaystyle -\dfrac23 \int_0^1 \left( \dfrac{x}{x^2+1} - \dfrac2{x^2+1} + \dfrac1{x+1} \right) \, dx \\ \displaystyle &= \displaystyle -\dfrac23 \left [ \dfrac12 \ln |x^2+1| - 2 \tan^{-1}(x) + \ln |x+1| \right ]_0^1 \\ \displaystyle &= \displaystyle -\dfrac23 \left( \dfrac32 \ln 2 - \dfrac\pi 2 \right) \\ &= \dfrac\pi3 - \ln 2 \end{array}

This means A = 1 , B = 3 , C = 2 ⇒ A + B + C = 6 A = 1, B = 3, C= 2\Rightarrow A+B+C=\boxed6 .

6 Helpful 0 Interesting 4 Brilliant 0 Confused

Can you please explain how you got the second step?

Nihar Mahajan - 5 years, 3 months ago

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Do you mean this one: = ∑ n = 0 ∞ 2 3 [ 1 4 n + 1 − 3 4 n + 3 + 2 4 n + 4 ] \displaystyle = \displaystyle \sum_{n=0}^\infty \dfrac23 \left [\dfrac1{4n+1} - \dfrac3{4n+3} + \dfrac2{4n+4} \right ] ? Just partial fractions .

Pi Han Goh - 5 years, 3 months ago

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Nope , the next one.

Nihar Mahajan - 5 years, 3 months ago

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@Nihar Mahajan – If you write the first few terms of 1 4 n + 1 − 3 4 n + 3 + 2 4 n + 4 \dfrac1{4n+1} - \dfrac{3}{4n+3} + \dfrac2{4n+4} , we get

1 − 3 3 + 2 4 , 1 5 − 3 7 + 2 8 , 1 9 − 3 11 + 2 12 , … 1 - \dfrac33 + \dfrac24, \dfrac15 - \dfrac37 + \dfrac28, \dfrac19 - \dfrac3{11} + \dfrac2{12} , \ldots

Without simplifying all these terms, each of these terms have 3 fractions.

Let's compare the first fractions in each term:

1 , 1 5 , 1 9 , 1 13 , … 1, \dfrac15, \dfrac19, \dfrac1{13} , \ldots

All of these fractions can be expressed as ∫ 0 1 1 d x , ∫ 0 1 x 4 d x , ∫ 0 1 x 9 d x , … \displaystyle \int_0^1 1 \,dx , \int_0^1 x^4 \,dx , \int_0^1 x^9 \,dx , \ldots

Taking the sum of all these "first fractions" alone gives us

∫ 0 1 ( 1 + x 4 + x 8 + ⋯ ) d x \displaystyle \int_0^1 (1 + x^4 + x^8 + \cdots ) \, dx

The integrand is an infinite geometric progression sum , with first term 1 and common ratio x 4 x^4 , this allows us to simplify the integral to ∫ 0 1 1 1 − x 4 d x \displaystyle \int_0^1 \dfrac 1{1-x^4} \, dx .

Similarly, by comparing the second fractions in each terms and taking their sum gives us an integrand of another infinite geometric progression sum with first term 3 x 2 3x^2 and common ratio x 4 x^4 , thus the sum is ∫ 0 1 3 x 2 1 − x 4 d x \displaystyle \int_0^1 \dfrac{3x^2}{1-x^4} \, dx .

And of course, we can work out the third fractions as well to obtain the sum of ∫ 0 1 2 x 3 1 − x 4 d x \displaystyle \int_0^1 \dfrac{2x^3}{1-x^4} \, dx .

Combining these three integrals gives us the following line.

Pi Han Goh - 5 years, 3 months ago

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@Pi Han Goh – Thank you!!!

Nihar Mahajan - 5 years, 3 months ago

You rarely post solutions in calculus. But when u post it is really elegant. Elegance over 9000!

Aditya Kumar - 5 years, 3 months ago

Here's an overkill approach:

Let f ( x ) = ∑ n = 0 ∞ 4 ( 4 n + 1 ) ( 4 n + 3 ) ( 4 n + 4 ) x 4 n + 4 \displaystyle f(x) = \sum_{n=0}^\infty \dfrac{ 4}{(4n+1)(4n+3)(4n+4)} x^{4n+4} , so we want to find f ( 1 ) f(1) . Differentiating it a couple of times yields

f ( 4 ) ( x ) = ∑ n = 0 ∞ 4 ( 4 n + 1 ) ( 4 n + 2 ) ( 4 n + 3 ) ( 4 n + 4 ) ( 4 n + 1 ) ( 4 n + 3 ) ( 4 n + 4 ) x 4 n = ∑ n = 0 ∞ ( 16 n + 8 ) x 4 n = 16 ∑ n = 0 ∞ n x 4 n + 8 ∑ n = 0 ∞ x 4 n , an infinite AGP sum and an infinite GP sum = 16 x 4 ( 1 − x 4 ) 2 + 8 1 − x 4 = 8 + 8 x 4 ( 1 − x 4 ) 2 \begin{aligned} f^{(4)}(x) &=& \sum_{n=0}^\infty \dfrac{4\cancel{(4n+1)}(4n+2)\cancel{(4n+3)}\cancel{(4n+4)}} {\cancel{(4n+1)}\cancel{(4n+3)}\cancel{(4n+4)}} x^{4n} \\ &=& \sum_{n=0}^\infty (16n+8)x^{4n} \\ &=& 16\sum_{n=0}^\infty n x^{4n}+8 \sum_{n=0}^\infty x^{4n}, \quad \text{an infinite AGP sum and an infinite GP sum} \\ &=& \dfrac{16x^4}{(1-x^4)^2} + \dfrac8{1-x^4} = \dfrac{8+8x^4}{(1-x^4)^2} \end{aligned}

Integrating back gives us

f ( 1 ) = ∫ 0 1 ∫ 0 z ∫ 0 y ∫ 0 w 8 + 8 v 4 ( 1 − v 4 ) 2 d v d w d y d z = ∫ 0 1 ∫ 0 z ∫ 0 y ( − 4 w w 4 − 1 − ln ⁡ ∣ 1 − w ∣ + ln ⁡ ∣ w + 1 ∣ + 2 tan ⁡ − 1 w ) d w d y d z = ∫ 0 1 ∫ 0 z ( − ln ⁡ ( 1 − y 2 ) − ( y − 1 ) ln ⁡ ∣ 1 − y ∣ + y ln ⁡ ∣ 1 + y ∣ + y ln ⁡ ∣ y + 1 ∣ + ln ⁡ ∣ y + 1 ∣ + 2 y tan ⁡ − 1 y ) d y d z = ∫ 0 1 1 2 ( z 2 ln ⁡ ∣ z + 1 ∣ − 2 z ln ⁡ ∣ 1 − z 2 ∣ + ( − z 2 + 2 z + 1 ) ln ⁡ ∣ 1 − z ∣ + 2 ( z 2 + 1 ) tan ⁡ − 1 ( z ) + 2 x ln ⁡ ∣ z + 1 ∣ − ln ⁡ ∣ z + 1 ∣ ) d z = π 3 − ln ⁡ 2 \begin{aligned} f(1) & = & \int_0^1 \int_0^z \int_0^y \int_0^w \dfrac{8+8v^4}{(1-v^4)^2} \, dv\; dw\; dy\; dz \\ & = & \int_0^1 \int_0^z \int_0^y \left( -\dfrac{4w}{w^4 - 1} - \ln|1 - w| + \ln |w + 1| + 2\tan^{-1} w \right) \, dw\; dy\; dz \\ & = & \int_0^1 \int_0^z \left( -\ln(1-y^2) - (y-1)\ln|1-y| + y \ln|1+y| + y \ln|y+1| + \ln|y+1| + 2y\tan^{-1} y \right) \, dy\; dz \\ & = & \int_0^1 \dfrac12 \left(z^2 \ln |z+1 | -2 z \ln | 1-z^2|+(-z^2+2 z+1) \ln|1-z|+2 (z^2+1) \tan^{-1}(z)+2 x \ln|z+1|-\ln|z+1| \right) \, dz \\ & = & \dfrac{\pi}3 - \ln 2 \end{aligned}

Note that I've skipped out a lot of integration steps which are all extremely tedious to evaluate. You should only be doing this method if you really have nothing else better to do or have no life .

Pi Han Goh - 5 years, 3 months ago

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Wow! Nice approach! I will try that out!

Aditya Kumar - 5 years, 3 months ago

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