Not easy as sum of squares -2

Algebra Level 5

∑ k = 1 ∞ 1 k 2 = π 2 6 , S i = ∑ k = 1 ∞ i ( 36 k 2 − 1 ) i \sum_{ k=1 }^\infty{ \frac { 1 }{ { k }^{ 2 } } =\frac { { \pi }^{ 2 } }{ 6 } },\qquad { S }_{ i }=\sum_{ k=1 }^\infty\frac { i }{ { \big(36{ k }^{ 2 }-1\big) }^{ i } }

Given the above, S 1 + S 2 S_{1}+S_{2} can be represented as π 2 a − b c \dfrac { { \pi }^{ 2 } }{ a } -\dfrac { b }{ c } .

Find the value of a + b + c a+b+c with c c a prime number.


The answer is 21.

View wiki
Gautam Sharma by Gautam Sharma , 23, India

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.

3 solutions

Wei Xian Lim
Feb 7, 2015

S 2 = ∑ k = 1 ∞ 2 ( 36 k 2 − 1 ) 2 S_{2}=\sum\limits_{k=1}^{\infty}\frac{2}{(36k^2-1)^2}

= 1 2 ∑ k = 1 ∞ ( 1 6 k − 1 − 1 6 k + 1 ) 2 \quad\;=\frac{1}{2}\sum\limits_{k=1}^{\infty}\big(\frac{1}{6k-1}-\frac{1}{6k+1}\big)^2

= 1 2 ∑ k = 1 ∞ [ ( 1 6 k − 1 ) 2 + ( 1 6 k + 1 ) 2 ] − ∑ k = 1 ∞ 1 ( 6 k − 1 ) ( 6 k + 1 ) \quad\;=\frac{1}{2}\sum\limits_{k=1}^{\infty}\big[\big(\frac{1}{6k-1}\big)^2+\big(\frac{1}{6k+1}\big)^2\big]-\sum\limits_{k=1}^{\infty}\frac{1}{(6k-1)(6k+1)}

= 1 2 ∑ k = 1 ∞ [ ( 1 6 k − 1 ) 2 + ( 1 6 k + 1 ) 2 ] − ∑ k = 1 ∞ 1 36 k 2 − 1 \quad\;=\frac{1}{2}\sum\limits_{k=1}^{\infty}\big[\big(\frac{1}{6k-1}\big)^2+\big(\frac{1}{6k+1}\big)^2\big]-\sum\limits_{k=1}^{\infty}\frac{1}{36k^2-1}

S 1 + S 2 = 1 2 ∑ k = 1 ∞ [ ( 1 6 k − 1 ) 2 + ( 1 6 k + 1 ) 2 ] S_{1}+S_{2}=\frac{1}{2}\sum\limits_{k=1}^{\infty}\big[\big(\frac{1}{6k-1}\big)^2+\big(\frac{1}{6k+1}\big)^2\big]

= 1 2 [ ∑ k = 1 ∞ ( 1 2 k − 1 ) 2 − ∑ k = 1 ∞ ( 1 3 ( 2 k − 1 ) ) 2 − 1 ] \qquad\quad\;\,\,=\frac{1}{2}\big[\sum\limits_{k=1}^{\infty}\big(\frac{1}{2k-1}\big)^2-\sum\limits_{k=1}^{\infty}\big(\frac{1}{3(2k-1)}\big)^2-1\big]

= 4 9 ∑ k = 1 ∞ ( 1 2 k − 1 ) 2 − 1 2 \qquad\quad\;\,\,=\frac{4}{9}\sum\limits_{k=1}^{\infty}\big(\frac{1}{2k-1}\big)^2-\frac{1}{2}

∑ k = 1 ∞ ( 1 2 k − 1 ) 2 = ∑ k = 1 ∞ 1 k 2 − ∑ k = 1 ∞ ( 1 2 k ) 2 \sum\limits_{k=1}^{\infty}\big(\frac{1}{2k-1}\big)^2=\sum\limits_{k=1}^{\infty}\frac{1}{k^2}-\sum\limits_{k=1}^{\infty}\big(\frac{1}{2k}\big)^2

= π 2 8 \qquad\qquad\quad=\frac{\pi^2}{8}

Hence, S 1 + S 2 = π 2 18 − 1 2 S_{1}+S_{2}=\frac{\pi^2}{18}-\frac{1}{2}

Answer: 18 + 1 + 2 = 21 18+1+2=\boxed{21}

20 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice solution!Keep up the good work

Gautam Sharma - 6 years, 4 months ago

This is what the question demands.perfect

Gautam Sharma - 6 years, 4 months ago

same method, and nicely explained!

Aareyan Manzoor - 6 years, 4 months ago

Good effort!

Lu Chee Ket - 6 years, 4 months ago

In the third to last line, how do you separate 1/(2k-1). Is this a series expansion?

Trevor Arashiro - 6 years, 3 months ago

Log in to reply

∑ n = 1 ∞ ( 1 ( 2 k − 1 ) 2 ) = 1 1 + 1 3 2 + 1 5 2 . . . \sum_{n=1}^\infty (\dfrac{1}{(2k-1)^2})=\dfrac{1}{1}+\dfrac{1}{3^2}+\dfrac{1}{5^2}... = 1 1 + 1 2 2 + 1 3 2 . . . − ( 1 2 2 + 1 4 2 . . . . ) =\dfrac{1}{1}+\dfrac{1}{2^2}+\dfrac{1}{3^2}...-(\dfrac{1}{2^2}+\dfrac{1}{4^2}....) = π 2 6 − 1 4 ( 1 1 + 1 2 2 + 1 3 2 . . . . . . ) =\dfrac{\pi^2}{6}-\dfrac{1}{4}(\dfrac{1}{1}+\dfrac{1}{2^2}+\dfrac{1}{3^2}......) = π 2 6 − π 2 24 = π 2 8 =\dfrac{\pi^2}{6}-\dfrac{\pi^2}{24}=\dfrac{\pi^2}{8}

Aareyan Manzoor - 6 years, 3 months ago

Log in to reply

OHhhhh, thank you, Now i get it. Very good explanation

Trevor Arashiro - 6 years, 3 months ago

No , ∑ k = 1 ∞ 1 ( 2 k − 1 ) 2 \sum _{ k=1 }^{ \infty }{ \frac { 1 }{ { (2k-1) }^{ 2 } } } is the sum of reciprocal squares of odd integers from 1 to ∞ \infty . And that is equal to sum of reciprocal squares of natural integers - sum of reciprocal squares of even natural no.s

Gautam Sharma - 6 years, 3 months ago
Joseph Miller
Feb 12, 2015

I must admit I did not immediately see a solution that did not use complex analysis, so here is a solution using complex analysis.

∑ k = − ∞ ∞ 1 36 k 2 − 1 = − [ R e s ( π cot ⁡ π k 36 k 2 − 1 , k = − 1 / 6 ) + R e s ( π cot ⁡ π k 36 k 2 − 1 , k = 1 / 6 ) ] = − π 2 3 , \displaystyle\sum_{k=-\infty}^{\infty}\frac{1}{36k^2-1}=-\left[Res\left(\frac{\pi \cot{\pi k}}{36k^2-1},k=-1/6\right) \\ +Res\left(\frac{\pi \cot{\pi k}}{36k^2-1},k=1/6\right)\right]\\ =-\frac{\pi}{2\sqrt3},

and

∑ k = − ∞ ∞ 2 ( 36 k 2 − 1 ) 2 = − [ R e s ( 2 π cot ⁡ π k ( 36 k 2 − 1 ) 2 , k = − 1 / 6 ) + R e s ( 2 π cot ⁡ π k ( 36 k 2 − 1 ) 2 , k = 1 / 6 ) ] = π 2 3 + π 2 9 . \displaystyle\sum_{k=-\infty}^\infty\frac{2}{(36k^2-1)^2}=-\left[Res\left(\frac{2\pi \cot{\pi k}}{(36k^2-1)^2},k=-1/6\right) \\ +Res\left(\frac{2\pi \cot{\pi k}}{(36k^2-1)^2},k=1/6\right)\right]\\ =\frac{\pi}{2\sqrt{3}}+\frac{\pi^2}{9}.

Thus,

2 S 1 − 1 = − π 2 3 2S_1-1=-\frac{\pi}{2\sqrt3} and

2 S 2 + 1 = π 2 3 + π 2 9 2S_2+1=\frac{\pi}{2\sqrt{3}}+\frac{\pi^2}{9} which yields

S 1 = 1 2 − π 4 3 S_1=\frac{1}{2}-\frac{\pi}{4\sqrt{3}} and

S 2 = − 1 + π 4 3 + π 2 18 S_2=-1+\frac{\pi}{4\sqrt{3}}+\frac{\pi^2}{18} .

Finally, we have

S 1 + S 2 = π 2 18 − 1 2 . S_1+S_2=\frac{\pi^2}{18}-\frac{1}{2}.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

I dont know what is it, but it seems to be very cool!

Gautam Sharma - 6 years, 3 months ago
Lu Chee Ket
Feb 9, 2015

S1 converged very fast while S2 converged more slowly. For S1 + S2:

| 0.048311355608886498 - 0.0483113556160755 | = 7.1889808E-12

0.048311355616075478824138388882008 = Pi^2/ 18 - 1/ 2

18 + 1 + 2 = 21

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...