Romanov's Series

Calculus Level 5

At first, I was amazed to those who could answer my recent OPs. I wonder, what kind of approach did they use to solve my OPs like:

  1. Possibly the strangest calculus problem??

  2. Possibly the strangest calculus problem?? (Part 2)

  3. Romanov's Integral

  4. Triple Cotangents Integral

But then I have a suspicion that they probably cheated since they answered them but posted no solutions. Besides, those problems can't be tackled by using softwares like Wolfram|Alpha, Mathematica, Maple, etc unless using special algorithms. Well, I hope I am wrong. (◠‿◠)

Anyway, here is a problem that I've just created.

∑ k = 1 ∞ ( − 1 ) k 2 k − 1 ( 2 k − 1 ) 4 − 16 = a π + b π s e c h c π d \large\sum_{k=1}^\infty (-1)^{k}\frac{2k-1}{(2k-1)^4-16}=\frac{\color{#D61F06}{a}\pi+\color{#D61F06}{b}\pi\,{\rm{sech}}\,\color{#D61F06}{c}\pi}{\color{#D61F06}{d}}

Compute a + b + c + d \large\, \color{#D61F06}{a}+\color{#D61F06}{b}+\color{#D61F06}{c}+\color{#D61F06}{d} .


The answer is 35.00.

Anastasiya Romanova by Anastasiya Romanova , 20, Russia

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1 solution

We take: x = 2 k − 1 x = 2k - 1

Then: x x 4 − 16 = x ( x − 2 ) ( x + 2 ) ( x 2 + 4 ) \frac{x}{x^4 - 16} = \frac{x}{(x-2)(x+2)(x^2 + 4)}

By Partial Fractions: x x 4 − 16 = A x − 2 + B x + 2 + C x + D x 2 + 4 \frac{x}{x^4 - 16} = \frac{A}{x-2} + \frac{B}{x+2} + \frac{Cx + D}{x^2 + 4} x = ( x + 2 ) ( x 2 + 4 ) A + ( x − 2 ) ( x 2 + 4 ) B + ( x − 2 ) ( x + 2 ) ( C x + D ) x = (x+2)(x^2 + 4)A + (x-2)(x^2+4)B + (x-2)(x+2)(Cx+D)

We take values for x x as {2, -2, 0, 1}

So, we get:

A = 1 16 =\frac{1}{16} , B = = 1 16 =\frac{1}{16} , D = 0 = 0 , C = − 1 8 =-\frac{1}{8}

Rewriting:

∑ k = 1 ∞ 1 16 ( − 1 ) k 2 k − 1 − 2 + 1 16 ( − 1 ) k 2 k − 1 + 2 − 1 8 ( − 1 ) k ( 2 k − 1 ) ( 2 k − 1 ) 2 + 4 \sum_{k=1}^{\infty} \frac{1}{16}\frac{(-1)^k}{2k - 1 - 2} + \frac{1}{16}\frac{(-1)^k}{2k - 1 + 2} - \frac{1}{8}\frac{(-1)^{k}(2k - 1)}{(2k - 1)^{2} + 4}

∑ k = 1 ∞ ( − 1 ) k 16 ( 1 2 k − 3 + 1 2 k + 1 ) + ∑ k = 1 ∞ − ( − 1 ) k 8 2 k − 1 ( 2 k − 1 ) 2 + 4 \sum_{k=1}^{\infty} \frac{(-1)^{k}}{16} ( \frac { 1 }{ 2k-3 } + \frac {1 }{ 2k+1} ) + \sum_{k=1}^{\infty} -\frac{(-1)^k}{8} \frac{2k-1}{(2k-1)^{2} + 4}

In the first summation:

= 1 16 ( − ( − 1 + 1 3 ) + ( 1 + 1 5 ) − ( 1 3 + 1 7 ) + . . . ) = \frac{1}{16} (-(-1 + \frac{1}{3}) + (1 + \frac{1}{5}) - (\frac{1}{3} + \frac{1}{7}) + ... )

= 1 16 ( 2 − 2 3 + 2 5 − 2 7 + 2 9 − 2 11 + . . . ) = \frac{1}{16}(2 - \frac{2}{3} + \frac{2}{5} - \frac{2}{7} + \frac{2}{9} - \frac{2}{11} + ... )

1 8 ∑ k = 1 ∞ ( − 1 ) k + 1 2 k − 1 = 1 8 π 4 \frac{1}{8} \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{2k-1} = \frac{1}{8} \frac{\pi}{4}

In the second summation:

− 1 8 ∑ k = 1 ∞ ( − 1 ) k ( 2 k − 1 ) ( 2 k − 1 ) 2 + 4 = − 1 8 − π s e c h ( π ) 4 -\frac{1}{8} \sum_{k=1}^{\infty} \frac{(-1)^k(2k-1)}{(2k-1)^{2} + 4} = -\frac{1}{8}\frac{-\pi sech(\pi)}{4}

Then:

π 32 + π s e c h ( π ) 32 = π + π s e c h ( π ) 32 \frac{\pi}{32} + \frac{\pi sech(\pi)}{32} = \frac{\pi + \pi sech(\pi)}{32}

So: a = 1 , b = 1 , c = 1 , d = 32 a = 1, b = 1, c = 1, d = 32

a + b + c + d = 35 a + b + c + d = \boxed{35}

10 Helpful 0 Interesting 0 Brilliant 0 Confused

Oh you meant this one. Nicely done! Very clever approach. ⌣ ¨ \ddot\smile

Anastasiya Romanova - 6 years, 6 months ago

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Nice problem :)

Joel Antonio Vásquez - 6 years, 6 months ago

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Thanks! I like user like you, first one who solves it, first one also who posts its solution :)

Anastasiya Romanova - 6 years, 6 months ago

I too considered the series - ∑ n = 1 ∞ ( − 1 ) n ( 2 n − 1 ) ( n − 1 / 2 ) 2 + x 2 = s e c h ( x ) \sum_{n=1}^{\infty}{\frac{{(-1)}^{n}(2n-1)}{{(n-1/2)}^{2} + {x}^{2}}} = sech(x) . But I did so 2 times, rather than splitting x 2 − 4 {x}^{2} - 4 into ( x − 2 ) ( x + 2 ) (x-2)(x+2) . You could have used the same series with the imaginary root in x x which will give that − - .

Kartik Sharma - 6 years, 4 months ago

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But your method is much smoother! +1'd! Nice problem! @Anastasiya Romanova

Kartik Sharma - 6 years, 4 months ago

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