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.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

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}

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...