Let's do some calculus! (37)

Calculus Level 5

∑ n = 0 ∞ cos ⁡ ( 3 π 5 ( 2 n + 1 ) ) + cos ⁡ ( π 5 ( 2 n + 1 ) ) ( 2 n + 1 ) 2 \sum_{n=0}^{\infty} \dfrac{ \cos \left( \frac{3 \pi}{5} \left(2n+1 \right) \right) + \cos \left( \frac{\pi}{5} \left(2n+1 \right) \right)}{{\left(2n+1 \right)}^2}

The infinite sum above can be represented as π a b \dfrac{{\pi}^a}{b} , where a a and b b are positive integers. Find the value of b a \dfrac ba .

This problem is inspired from Ishan Singh .

For more problems on calculus, click here .


The answer is 10.

Tapas Mazumdar by Tapas Mazumdar , 20, India

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1 solution

Tapas Mazumdar
Oct 14, 2016

This solution is somewhat of a plagiarism. @Chew-Seong Cheong posted a solution with this approach which I have used to solve this problem. You can see Chew's solution in this note . Sorry sir!

Consider cos ⁡ ( 3 π 5 ( 2 n + 1 ) ) + cos ⁡ ( π 5 ( 2 n + 1 ) ) \cos \bigg( \dfrac{3 \pi}{5} \left(2n+1 \right) \bigg) + \cos \bigg( \dfrac{\pi}{5} \left(2n+1 \right) \bigg) . For the first few values of n n ,

n = 0 : cos ⁡ 3 π 5 + cos ⁡ π 5 = 1 2 n = 1 : cos ⁡ 9 π 5 + cos ⁡ 3 π 5 = 1 2 n = 2 : cos ⁡ 3 π + cos ⁡ π = − 2 n = 3 : cos ⁡ 21 π 5 + cos ⁡ 7 π 5 = 1 2 n = 4 : cos ⁡ 27 π 5 + cos ⁡ 9 π 5 = 1 2 n = 5 : cos ⁡ 33 π 5 + cos ⁡ 11 π 5 = 1 2 n = 6 : cos ⁡ 39 π 5 + cos ⁡ 13 π 5 = 1 2 n = 7 : cos ⁡ 9 π + cos ⁡ 3 π = − 2 n = 8 : cos ⁡ 51 π 5 + cos ⁡ 17 π 5 = 1 2 . . . . . . . . . \begin{array}{c}\displaystyle n=0 &:& \cos \frac{3\pi}{5} + \cos \frac{\pi}{5} & = \frac 12 \\ \\ \displaystyle n=1 &:& \cos \frac{9\pi}{5} + \cos \frac{3\pi}{5} & = \frac 12 \\ \\ \displaystyle n=2 &:& \cos 3\pi + \cos \pi & = -2 \\ \\ \displaystyle n=3 &:& \cos \frac{21\pi}{5} + \cos \frac{7\pi}{5} & = \frac 12 \\ \\ \displaystyle n=4 &:& \cos \frac{27\pi}{5} + \cos \frac{9\pi}{5} & = \frac 12 \\ \\ \displaystyle n=5 &:& \cos \frac{33\pi}{5} + \cos \frac{11\pi}{5} & = \frac 12 \\ \\ \displaystyle n=6 &:& \cos \frac{39\pi}{5} + \cos \frac{13\pi}{5} & = \frac 12 \\ \\ \displaystyle n=7 &:& \cos 9\pi + \cos 3\pi & = -2 \\ \\ \displaystyle n=8 &:& \cos \frac{51\pi}{5} + \cos \frac{17\pi}{5} & = \frac 12 \\ ... & & ... & ... \end{array}

Thus,

cos ⁡ ( 3 π 5 ( 2 n + 1 ) ) + cos ⁡ ( π 5 ( 2 n + 1 ) ) = { 1 2 if n mod 5 ≠ 2 − 2 if n mod 5 = 2 \cos \bigg( \dfrac{3 \pi}{5} \left(2n+1 \right) \bigg) + \cos \bigg( \dfrac{\pi}{5} \left(2n+1 \right) \bigg) = \begin{cases} \frac 12 & \text{if } n \text{ mod }5 \neq 2 \\ -2 & \text{if } n \text{ mod }5 = 2 \end{cases}

Then we have,

S = ∑ n = 0 ∞ cos ⁡ ( 3 π 5 ( 2 n + 1 ) ) + cos ⁡ ( π 5 ( 2 n + 1 ) ) ( 2 n + 1 ) 2 = 1 2 1 2 + 1 2 3 2 − 2 5 2 + 1 2 7 2 + 1 2 9 2 + 1 2 11 2 + 1 2 13 2 − 2 15 2 + ⋯ = 1 2 1 2 + 1 2 3 2 + 1 2 5 2 − 5 2 5 2 + 1 2 7 2 + 1 2 9 2 + 1 2 11 2 + 1 2 13 2 + 1 2 1 5 2 − 5 2 15 2 + ⋯ = 1 2 ⋅ ∑ n = 0 ∞ 1 ( 2 n + 1 ) 2 − 5 2 ⋅ ∑ n = 0 ∞ 1 ( 10 n + 5 ) 2 = 1 2 ⋅ ∑ n = 0 ∞ 1 ( 2 n + 1 ) 2 − 1 10 ⋅ ∑ n = 0 ∞ 1 ( 2 n + 1 ) 2 = ( 1 2 − 1 10 ) ⋅ ∑ n = 0 ∞ 1 ( 2 n + 1 ) 2 = 2 5 ⋅ π 2 8 = π 2 20 \begin{aligned} S &= \sum_{n=0}^{\infty} \dfrac{ \cos \bigg( \dfrac{3 \pi}{5} \left(2n+1 \right) \bigg) + \cos \bigg( \dfrac{\pi}{5} \left(2n+1 \right) \bigg)}{{\left(2n+1 \right)}^2} \\ &= \dfrac{\frac 12}{1^2} + \dfrac{\frac 12}{3^2} - \dfrac{2}{5^2} + \dfrac{\frac 12}{7^2} + \dfrac{\frac 12}{9^2} + \dfrac{\frac 12}{{11}^2} + \dfrac{\frac 12}{{13}^2} - \dfrac{2}{{15}^2} + \cdots \\ &= \dfrac{\frac 12}{1^2} + \dfrac{\frac 12}{3^2} + \dfrac{\frac 12}{5^2} - \dfrac{\frac 52}{5^2} + \dfrac{\frac 12}{7^2} + \dfrac{\frac 12}{9^2} + \dfrac{\frac 12}{{11}^2} + \dfrac{\frac 12}{{13}^2} + \dfrac{\frac 12}{15^2} - \dfrac{\frac 52}{{15}^2} + \cdots \\ &= \dfrac 12 \cdot \sum_{n=0}^{\infty} \dfrac{1}{{\left(2n+1\right)}^2} - \dfrac 52 \cdot \sum_{n=0}^{\infty} \dfrac{1}{{\left(10n+5\right)}^2} \\ &= \dfrac 12 \cdot \sum_{n=0}^{\infty} \dfrac{1}{{\left(2n+1\right)}^2} - \dfrac{1}{10} \cdot \sum_{n=0}^{\infty} \dfrac{1}{{\left(2n+1\right)}^2} \\ &= \left( \dfrac 12 - \dfrac{1}{10} \right) \cdot \sum_{n=0}^{\infty} \dfrac{1}{{\left(2n+1\right)}^2} \\ &= \dfrac 25 \cdot \dfrac{{\pi}^2}{8} & \small \color{#3D99F6} \\ &= \dfrac {{\pi}^2}{20} \end{aligned}

( \bigg( See why ∑ n = 0 ∞ 1 ( 2 n + 1 ) 2 = π 2 8 \displaystyle \sum_{n=0}^{\infty} \dfrac{1}{{\left(2n+1\right)}^2} = \dfrac{{\pi}^2}{8} here ) \bigg)

Thus, a = 2 a=2 and b = 20 b=20 which gives us b a = 10 \dfrac ba = \boxed{10} .

3 Helpful 0 Interesting 0 Brilliant 0 Confused

It is quite okay, there is no copyright to solution.

Chew-Seong Cheong - 4 years, 8 months ago

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Thanks for not getting offended by this. :)

Tapas Mazumdar - 4 years, 8 months ago

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All you need to mention from whom the idea of solution has come from. Just like we mention Lagrange multipliers to give credit to Lagrange. How I wish I have my own theorem and method.

Chew-Seong Cheong - 4 years, 8 months ago

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@Chew-Seong Cheong – Haha. Just keep exploring. Who knows one day you are in the middle of a certain problem and you realize, "Eureka! I've just figured the most efficient way to solve this problem."

Hope your margin won't be as short as Fermat had so that you can enlighten us with your work as well rather than providing us a note saying, "I've discovered a marvelous proof that this narrow margin is too short to contain."

On another note, I had kept this thought in my mind before posting this problem that in the solution, I would provide credits to you. :)

Tapas Mazumdar - 4 years, 8 months ago

In general ∑ n = 0 ∞ ( ∑ k = 0 ⌊ p 2 ⌋ − 1 ( cos ⁡ ( π p ( 2 k + 1 ) ( 2 n + 1 ) ) ) ( 2 n + 1 ) s ) = ( 1 2 − 1 2 p s − ⌊ p 2 ⌋ p s ) ( 1 − 2 − s ) ζ ( s ) \displaystyle \sum_{n=0}^{\infty} \left(\frac {\displaystyle\sum_{k=0}^{\left \lfloor \frac p2 \right\rfloor -1} \left(\cos \left(\frac {\pi}{p}(2k+1)(2n+1)\right)\right) }{(2n+1)^s}\right) =\left(\frac 12 -\frac {1}{2p^s }-\frac {\left\lfloor \frac p2\right\rfloor}{p^s}\right) (1-2^{-s})\zeta(s)

Where p p is a prime number, s ∈ R s\in R and s > 1 s\gt 1

Also ⌊ . ⌋ \lfloor. \rfloor denotes the floor function while ζ ( s ) \zeta(s) denotes the Riemann Zeta function.

Rohan Shinde - 2 years, 2 months ago

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I need some help with the proof. Can you please provide some hint?

Tapas Mazumdar - 1 year, 8 months ago

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@Tapas Mazumdar You can try to convert it into a finite sum of Polylogarithms and then use the properties of Polylogarithms and the multiplication formula of polylogarithms.

Rohan Shinde - 1 year, 8 months ago

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