So Many Sequences (Weekly Brain Potion Discussion)
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.
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=1infinity(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.
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...