FoxTrot

Calculus Level 5

k = 1 ( 1 ) k + 1 k 2 k 3 + 1 = 1 A [ 1 ln ( B ) + π sech ( 3 π C ) ] \large \sum\limits_{k=1}^{\infty} \frac{(-1)^{k+1}k^2}{k^3+1}=\frac{1}{A}\left[1-\ln(B)+\pi \, \text{sech}\left(\frac{\sqrt{3}\:\pi}{C}\right)\right]

Find the value of A + B + C A+B+C , where A , B A,B and C C are positive integers.


The answer is 7.

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1 solution

Otto Bretscher
Jan 4, 2016

I would have been lost without the FoxTrot clue! Here is the relevant FoxTrot comic strip and a brief solution. (Sorry, I don't feel like retyping their fine solution.)

That's perfectly fine, I just happened to stumble across it and thought it was pretty cool.

I wish there was a comic dedicated to mathematics. I'd buy it...

Isaac Buckley - 5 years, 5 months ago

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xkcd and Saturday Morning Breakfast Cereal

Pi Han Goh - 5 years, 5 months ago

The solution provided in that link skips out a lot of major steps and it's not easily understood.

Pi Han Goh - 5 years, 5 months ago

Ohhh I got it! Write the summation as

k = 1 ( ( 2 k 1 ) 2 ( 2 k 1 ) 3 + 1 ( 2 k ) 2 ( 2 k ) 3 + 1 ) \sum_{k=1}^\infty \left( \dfrac{(2k-1)^2}{(2k-1)^3 + 1} - \dfrac{(2k)^2}{(2k)^3 + 1} \right)

Then apply partial fractions decomposition.

Unfortunately, that still requires a lot of work...

Pi Han Goh - 5 years, 2 months ago

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