Limit + Summation = Integration (3)

Calculus Level 5

lim ⁡ n → ∞ ∑ k = 1 n n 3 n 4 + k 4 = π + ln ⁡ ( a + b b ) c b \large \lim_{n\rightarrow\infty}\sum_{k=1}^{n}\frac{n^3}{n^4+k^4} = \frac{\pi+\ln{(a+b\sqrt{b})}}{c\sqrt{b}}

If a a , b b and c c are positive integers, find a + b + c a+b+c .


The answer is 9.

Tommy Li by Tommy Li , 22, Hong Kong

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1 solution

Chew-Seong Cheong
Jul 27, 2017

L = lim ⁡ n → ∞ ∑ k = 1 n n 3 n 4 + k 4 Divide up and down by n 4 = lim ⁡ n → ∞ ∑ k = 1 n 1 n 1 + ( k n ) 4 By Riemann sums = ∫ 0 1 1 1 + x 4 d x = ∫ 0 1 1 ( x 2 + 1 ) 2 − 2 x 2 d x = ∫ 0 1 1 ( x 2 + 2 x + 1 ) ( x 2 − 2 x + 1 ) d x By partial fractions \begin{aligned} L & = \lim_{n \to \infty} \sum_{k=1}^n \frac {n^3}{n^4+k^4} & \small \color{#3D99F6} \text{Divide up and down by }n^4 \\ & = \lim_{n \to \infty} \sum_{k=1}^n \frac {\frac 1n}{1+\left(\frac kn\right)^4} & \small \color{#3D99F6} \text{By Riemann sums} \\ & = \int_0^1 \frac 1{1+x^4} \ dx \\ & = \int_0^1 \frac 1{(x^2+1)^2-2x^2} \ dx \\ & = \int_0^1 \frac 1{(x^2+\sqrt2x+1)(x^2-\sqrt2x+1)} \ dx & \small \color{#3D99F6} \text{By partial fractions} \end{aligned}

= 1 4 ∫ 0 1 ( 2 x + 2 x 2 + 2 x + 1 − 2 x − 2 x 2 − 2 x + 1 ) d x = 1 4 ∫ 0 1 ( 1 2 ( 2 x + 2 ) + 1 x 2 + 2 x + 1 − 1 2 ( 2 x − 2 ) − 1 x 2 − 2 x + 1 ) d x = 1 4 ∫ 0 1 ( 1 2 ( 2 x + 2 ) x 2 + 2 x + 1 + 1 ( x + 1 2 ) 2 + 1 2 − 1 2 ( 2 x − 2 ) x 2 − 2 x + 1 + 1 ( x − 1 2 ) 2 + 1 2 ) d x = 1 4 ∫ 0 1 ( 1 2 ( 2 x + 2 ) x 2 + 2 x + 1 + 2 ( 2 x + 1 ) 2 + 1 − 1 2 ( 2 x − 2 ) x 2 − 2 x + 1 + 2 ( 2 x − 1 ) 2 + 1 ) d x = 1 4 [ 1 2 ln ⁡ ( x 2 + 2 x + 1 ) + 2 2 tan ⁡ − 1 ( 2 x + 1 ) − 1 2 ln ⁡ ( x 2 − 2 x + 1 ) + 2 2 tan ⁡ − 1 ( 2 x − 1 ) ] 0 1 = 1 4 2 [ ln ⁡ ( 2 + 2 ) + 2 ( 3 π 8 − π 4 ) − ln ⁡ ( 2 − 2 ) + 2 ( π 8 + π 4 ) ] = 1 4 2 [ ln ⁡ ( 2 + 2 2 − 2 ) + π ] = 1 4 2 [ π + ln ⁡ ( ( 2 + 2 ) 2 ( 2 − 2 ) ( 2 + 2 ) ) ] = 1 4 2 [ π + ln ⁡ ( 6 + 4 2 4 − 2 ) ] = π + ln ⁡ ( 3 + 2 2 ) 4 2 \begin{aligned} \ \ & = \frac 14 \int_0^1 \left(\frac {\sqrt 2 x + 2}{x^2+\sqrt2x+1} - \frac {\sqrt 2 x - 2}{x^2-\sqrt2x+1} \right) dx \\ & = \frac 14 \int_0^1 \left(\frac {\frac 1{\sqrt 2}(2x + \sqrt 2) + 1}{x^2+\sqrt2x+1} - \frac {\frac 1{\sqrt 2}(2x - \sqrt 2) - 1}{x^2-\sqrt2x+1} \right) dx \\ & = \frac 14 \int_0^1 \left(\frac {\frac 1{\sqrt 2}(2x + \sqrt 2)}{x^2+\sqrt2x+1} + \frac 1{\left(x+\frac 1{\sqrt 2}\right)^2+\frac 12} - \frac {\frac 1{\sqrt 2}(2x - \sqrt 2)}{x^2-\sqrt2x+1} + \frac 1{\left(x-\frac 1{\sqrt 2}\right)^2+\frac 12} \right) dx \\ & = \frac 14 \int_0^1 \left(\frac {\frac 1{\sqrt 2}(2x + \sqrt 2)}{x^2+\sqrt2x+1} + \frac 2{(\sqrt 2 x + 1)^2+1} - \frac {\frac 1{\sqrt 2}(2x - \sqrt 2)}{x^2-\sqrt2x+1} + \frac 2{(\sqrt 2 x - 1)^2+1} \right) dx \\ & = \frac 14 \left[ \frac 1{\sqrt2} \ln (x^2+\sqrt2x+1) + \frac 2{\sqrt2} \tan^{-1}(\sqrt 2 x + 1) - \frac 1{\sqrt2} \ln (x^2-\sqrt 2 x+1) + \frac 2{\sqrt2} \tan^{-1}(\sqrt 2 x - 1) \right ]_0^1 \\ & = \frac 1{4\sqrt 2} \left[\ln (2+\sqrt2) + 2 \left(\frac {3\pi}8 - \frac \pi 4\right) - \ln (2-\sqrt2) + 2 \left(\frac \pi 8 + \frac \pi 4\right) \right] \\ & = \frac 1{4\sqrt 2} \left[\ln \left( \frac {2+\sqrt2}{2-\sqrt2}\right) + \pi \right] \\ & = \frac 1{4\sqrt 2} \left[\pi + \ln \left( \frac {(2+\sqrt2)^2}{(2-\sqrt2)(2+\sqrt2)}\right) \right] \\ & = \frac 1{4\sqrt 2} \left[\pi + \ln \left( \frac {6+4\sqrt2}{4-2}\right) \right] \\ & = \frac {\pi + \ln (3+2\sqrt 2)}{4\sqrt 2} \end{aligned}

⟹ a + b + c = 3 + 2 + 4 = 9 \implies a+b+c = 3+2+4 = \boxed{9}

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