Zoom International Math League: So Many Sequences (Weekly Brain Potion Discussion)

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So Many Sequences (Weekly Brain Potion Discussion)

For this week's brain potion we will explore some various problems involving sequences. Caution, you'll probably want a calculator for some of these!

For each of the sequences below: (i) Describe the pattern and calculate the first few numbers. (ii) What number do you think the sequence is approaching?

$$\displaystyle \frac{4}{1}, \frac{4}{1} - \frac{4}{3}, \frac{4}{1} - \frac{4}{3} + \frac{4}{5}, \frac{4}{1} - \frac{4}{3} + \frac{4}{5} - \frac{4}{7}, \ldots$$
$$\displaystyle 2\times \frac{2}{1}, 2\times \frac{2}{1}\frac{2}{3}, 2\times \frac{2}{1}\frac{2}{3}\frac{4}{3}, 2\times \frac{2}{1}\frac{2}{3}\frac{4}{3}\frac{4}{5}, 2\times \frac{2}{1}\frac{2}{3}\frac{4}{3}\frac{4}{5}\frac{6}{5},\ldots$$
$$3, 3 + \frac{4}{2\cdot3\cdot4}, 3 + \frac{4}{2\cdot3\cdot4} - \frac{4}{4\cdot5\cdot6}, 3 + \frac{4}{2\cdot3\cdot4} - \frac{4}{4\cdot5\cdot6} + \frac{4}{6\cdot7\cdot8}, \ldots$$
$$3, 3 + 1^2, 3 + \frac{1^2}{6 + 3^2}, 3 + \frac{1^2}{6 + \dfrac{3^2}{6+5^2}}, 3 + \frac{1^2}{6 + \dfrac{3^2}{6+\dfrac{5^2}{6+7^2}}}, \ldots$$

As a challenge, use Excel/Google sheets or write a program to help calculate the values of these sequences!

Please share any thoughts or questions you have below. If you have any other favorite sequences/patterns (bonus points for sticking with the "theme" of the sequences above) feel free to share them as well!

The first sequence modifies the previous term by adding (-1)^k-1(4/(2k-1)) to the previous term (need to think of a general formula). A₁=4, A₂=8/3, A₃=52/15...

The number that the sequence approaches is the answer to the infinite series ∑^∞_k=1(-1)^k-1(4/(2k-1)) which by the alternating series test has to converge (lim as k approaches infinity of (4/(2k-1)) is 0, and it decreases). Specifically, it approaches... π, actually, in accordance with the Wolfram Alpha widget (this is probably some other definition of π, actually, using infinite series and a touch of calculus.

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The second sequence required a bit more thinking, but I think I got it. Somewhat. I had to actually take out the sequence in order to get something comprehensive, but this will help later on. Basically, I noticed that denominators were the main problem here, so I tried to get rid of as many redundant denominators as I could, being left with terms such as (6*8)/(7²), for instance. I’ll call the new sequence b. Afterwards, I constructed the Pi notation: 4(which is b(1)) times ∏_n=1^infinity(2n)(2n-2)/(2n-1)²) to determine the value it approaches. The product approaches pi/4, so by multiplying it by 4, b(infinity) is pi.

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And if I wasn’t clear about the pattern of the sequence, it’s that the previous term is multiplied by (2n)(2n-2)/(2n-1)²

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Likewise, sequence 3 isn’t so hard. For each term, the sequence adds (-1)^k(4)/[(2n)(2n-1)(2n-2).] To determine the value the sequence approaches, start by ignore the 3 and just focus on the alternating values The value it approaches is 3 + the value of the infinite series ∑^∞_k=2(-1)^k(4)/[(2k-2)(2k-1)(2k)] which is π-3+3 or π. I’m starting to think these all converge to π as a sort of belated π day celebration. Incoming post where the infinite sum/products are used instead of π...

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My guy Darren finding the solution to the mathematical potion.

All of these can be rewritten as formulas and calculated via internet.

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Sequence 4 probably also converges to pi...

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