In class we have been toying with the idea of classifying diverging infinite series, such as the sum: Σ k (k = 1,∞) = 1 + 2 + 3 + … which, as we add it up, continues on to infinity. We also messed around with some series notations and came to a conclusion that the sum adds up to -1/12. Now, I have no intention to claim, nor prove, that it equals -1/12. In fact, I would like to do the opposite; I would like to show that it in no way converges. If an infinite series **converges**, that means it sums up to a real number s: Σ ak = s. Likewise, an infinite series **diverges** if the sum of the series equals ±∞ or it does not add to any value. There are a few methods I will use to prove that this series diverges, and these methods can also be used to determine whether any infinite series converges or diverges.

The first test I want to look at is the Root Test. In this test, we take the sequence a_{k} raise it to the power 1/k, and take the limit of that as k→∞. If this limit is greater than 1, the series diverges; if the limit is less than 1, it converges. For the sum of all integers, that gives us limit as k→∞ of k^{(1/k)} . As k→∞, limit of k^{(1/k) } → 1. Unfortunately the limit equals 1, which means it is inconclusive; there is a chance that this does indeed sum up to -1/12. Let’s leave the score at 0-0.

I will now put the series through the Ratio Test. It is called the Ratio Test because we take a ratio of the k+1 term divided by the k-th term and take the limit of that as k→∞: the limit of a_{k+1}/a_{k}. Similar to the Root Test, if the limit is greater than one it diverges, and if the limit is less than 1 it converges. For the dubious series which is under examination, that limit (k+1)/k, and as k→∞ the limit equals 1. Once again, this test is inconclusive. The score is still 0-0.

Now it’s time to get serious; so far every test I’ve done has been disappointingly empty of an answer. In this test if a series b_{k }> a_{k} and b_{k} converges, then a_{k} must also converge and if b_{k }< a_{k} and b_{k} diverges, then a_{k} diverges as well. My claim is that Σ k (k = 1,∞) diverges and now I will compare it with a series Σ 1 (k = 1,n) = n. We can see that 1+1+1+… grows much more slowly than 1+2+3+… thus we can use it as a comparison. By another test, the Term Test, if the series converges then lim a_{n }= 0. The series must diverge then if the limit doesn’t equal 0. The limit of 1 as n→∞ = 1 ≠ 0, thus the sum of 1 on (k=1,∞) diverges. Since Σ 1 (k = 1,∞) < Σ k (k = 1,∞), the latter series must also diverge. Now the score is 1 for divergence, and 0 for -1/12.

While it can now be seen that Σ k (k = 1,∞) does indeed diverge, the comparison test relies on having knowledge that a similar infinite series will either converge or diverge. In this case, I have to know that Σ 1 (k = 1,∞) diverges before I can compare it to the original series. I used the Term Test to show that Σ 1 (k = 1,∞) diverged, but I can also use it to show that Σ k (k = 1,∞) diverges. The limit as n→∞ of k = ∞, and by the Term Test diverges. The problem with the Term Test is that the limit a_{k} can be equal to zero, but the series can still diverge. Therefore, this test is only useful if the limit does not equal 0 or to insure that a converging series does indeed converge.

By both the Term and Comparison tests, I was able to show that Σ k (k = 1,∞) diverges and is not equal to -1/12. In class though, we determined that the infinite series itself doesn’t sum to a negative number, there is no possible way that adding large positive whole numbers together would result in a negative rational number, but rather -1/12 represented a value or categorization of the infinite sum Σ k (k = 1,∞). How did this number come about? The idea is definitely not of “fringe mathematics” and has some excellent arguments. The idea stems from Σ (-1)^{k} (k = 0,∞) = 1 – 1 + 1 – 1 + … The mathematician Srinivasa Ramanujan gave this sum a “value” of ½, since it seems to jump between one and zero equally, with ½ as the average.

The second sum Σ k(-1)^{k-1} (k = 1,∞) = ¼ as this sum added to itself equals Σ (-1)^{k} (k = 0,∞). That is 1 – (2 – 1) + (3 – 2) + … = 1 – 1 + 1 – 1 + … Finally, Σ k (k = 1,∞) – 4(Σ k (k = 1,∞)) = 1 + (2 – 4) + 3 – (4 – 8) + … equals 1 – 2 + 3 – 4 + … = Σ k(-1)^{k-1} (k = 1,∞), or rather -3(Σ k (k = 1,∞)) = ¼ → Σ k (k = 1,∞) = -1/12.

The specific stipulation given is that the series must go to its limit. The partial summations of any of these series will produce a number unlike the total summation. For the majority of mathematics, to say Σ k (k = 1,∞) = ∞ makes more sense and requires significantly less head-scratching than -1/12.

There is one more test for convergence that I did not talk about, as the infinite series I was examining did not apply, and that is the Integral Test. For Σ a_{k}, the function f(k) = a_{k}. The reason why I could not apply it is because it only works with positive, non-increasing functions bounded on [1,∞). If the integral of f(x) is a real number, then the series converges, whereas if that same integral diverges, then the series diverges. The first case I will look at is Σ 1/k (k = 1,∞), where f(x) = 1/x and is a positive, non-increasing function. The integral of 1/x = log(x) evaluated from 1 to ∞. This integral turns out to diverge, and therefore the series also diverges. The second case is similar: Σ 1/k^{2} (k = 1,∞), where f(x) = x^{-2}. The integral of 1/x^{2} = -(1/x) evaluated over the same interval. The sum of these two bounds is 0 – (-1) = 0 + 1 = 1. Since the indefinite integral converges, the series also converges.

These tests help determine whether a series converges or not. I used them to prove with basic mathematics that Σ k (k = 1,∞) diverges, rather than converging on a negative rational number. While the values given to divergent indefinite series can provide an idea of how they relate to each other, they require a fair amount of assumption and a lot of counterintuitive work to calculate. It is far easier and more practical to state that Σ k (k = 1,∞) diverges.

Sources:

http://www.lemiller.net/media/classfiles/notes.pdf (Foundations of Analysis by Joseph L. Taylor)

https://www.math.hmc.edu/calculus/tutorials/convergence/

http://mathworld.wolfram.com/RatioTest.html

http://en.wikipedia.org/wiki/Series_%28mathematics%29

http://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF