1. Alex Shen Dr. Blozy
PD 6 and 7
Math Major Writing Assignment Mini Research Part II Due: Nov. 5, 2015
I. Statement of the ResearcH Question
Does 1+2+3+4+5+... have a finite numerical value?
II. Origin of the Question
Recently, we began the concepts of Number Theory by using the method of
induction. For a practice question, we used the pattern of
1+2+3+4+5+...n=((1+n)n)/2. In this situation, there is a finite solution; it is clear that
this must be true for all the numbers in a finite series. That got me thinking about
going beyond a constraint of numbers. What if I went up to infinity? Would there be
no such answer, simply infinity? Or would it also follow some rule and I end up
getting a finite number?
Each number that I am adding up is increasing by one each time, so it is an
arithematic sequence, which I learned last year in Trig/Algebra II. There is a constant
difference between any two consecutive numbers. Now, I feel like this conundrum
might be a trick question that might end up diverging into two distinct answers.
Initially, I claimthat the answer would tend toward infinity.
This concept reminded me of one of the math problems we did in the
beginning of the year: √20+ √20+√20+... which ended up with a finite solution of 5.
Something that is going on infinitely had a solution that was finite! If that was true
for this particular example, it could also infer that in other infinite series of numbers,
it might work as well.
III. Investigation of the Question
There is a misconception to think that in mathematics, it is a linear process
where we are only doing things that are legitimate. However, I will view this
problem in a different approach by using partial sums of an infinite series or the sum
of a finite series of number of consecutive terms beginning with the first term.
Let’s start off with viewing this concept with three different infinite series of
numbers S1, S2, and our main series, S. (Note: All these series correlate with one
another. It’s just a heads up.)

2. Let S1= 1-1+1-1+1-1+1-1+...
How do we find the answer for this? Well, let’s start by putting parentheses
around two consecutive numbers:
S1= (1-1)+(1-1)+(1-1)+...
And if we simplify what is in the parentheses, we end up adding zeros.
Therefore, our answer must be 0...right?
What if we put the parentheses in a different location, say, starting with the
second term instead of the first:
S1= 1+(-1+1)+(-1+1)+(-1+1)+...
When we simplify this, we end up adding the first 1 with everything else
which are a whole lot of zeros. Therefore, the answer must be 1.
How can we have two answers? If we stop at an odd point, we get the
answer as 1:
1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 +...
If we stop at an even point, we get the answer as 0:
1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 +...
Therefore, by taking the average of the two possible solutions, we end up
getting ½ as our final answer for this infinite series of numbers.
I will confirm this in a different approach. Knowing that the two possible
answers ( 0 and 1), we will be trying to find infinite series, call it M. If we subtract M
from 1, it would equal to 1- (1-1+1-1+1-1+...). Simplifying that, we end up getting 1-
1+1-1+1-1+1-1+... or basically what we started with, M. Therefore, 1-M=M and by
bringing M over to one side, M=1/2.
IT IS OFFICIAL, S1= 1/2. Now, let’s go on to the next infinite series of
numbers, S2.
Let S2= 1-2+3-4+5-6+...
I will now add another copy of the same series, which would then give me:
S2= 1-2+3-4+5-6+...
+ S2= 1-2+3-4+5-6+...
____________________
2S2= 1-1+1-1+1-1+...
On the right side of the equation, we already solved for that infinte series of
numbers (S1) in the previous series!
Therefore, 2S2=1/2 and if we isolate S2, we have:
S2= ¼.
However,whenIaddthe two series
together,Iwill shiftthe bottomseries
slightlytothe right.Youwill see why
thisissignificantlater.

3. Finally, let’s look at our main infinite series that we are trying to solve, S.
Let S= 1+2+3+4+5+6+...
We will use S2 to aid us with our procedure because they have similar terms that
benefit in this situation:
S= 1+2+3+4+5+6+...
S2= 1-2+3-4+5-6+...
Let’s try and subtract S2 from S:
S= 1+2+3+4+5+6+...
- S2= 1-2+3-4+5-6+...
_____________________
S-S2= 1+2+3+4+5+6+...
-[1-2+3-4+5-6+...]
______________________
S-S2= 0+4+0+8+0+12+0+16+...
This means that S-S2= 4+8+12+16+...
If we factor out a 4 it would then equal to:
S-S2= 4(1+2+3+4+...)
In the parentheses, that is basically what we initally started with, the inifinte series,
S. Therefore:
S-S2= 4(S)
Furthermore, we already know the value of S2 which is 1/4, so we could subsitute
that into the equation:
S- (1/4)= 4S
-1/4= 3S
S= -1/12. !!!!!!!!!! WHAT?!?!?!?!

4. IV. Conclusion
From all that tedious work, I have finally concluded that an infinite series of
1+2+3+4+5+6+... is equal to -1/12. I speculated the answer to be infinity because by
adding up only positive numbers, I expect the answer to be infinity. However, from
the work that I have just done, it is a negative fraction of -1/12. It is baffling me, how
adding up positive integers would actually result to a negative fraction.
The procedure of coming up with the final answer does overrule my original
claimthat it the series of numbers from one to infinity would add up to be, well,
infinity. It is convincing how just by using some algebra, I was able to achieve
solutions that are mind boggling. Overall, this experience was unforgettable and not
what I expected it to be.
V. Extensions
Instead of adding up the numbers, we could try and multiply them instead.
We could also raise each number to an exponent and add all of them up from one to
infinity. Or rather than adding all the positive numbers, add all the negative numbers
(or simply subtraction) to negative infinity. By trying and using different
mathematical operations that would slightly tweak the original problem, the end
result could be totally unexpected and different from other series of infinite
numbers.
By replacing a series by its “regularized” values, each of the infinite series
contain a very significant and valuable number. Everything else, therefore, is
unimportant and we could just disregard it or “throw it away.”
VI. Reference
- Grandi’s Series: Quadratura circula et hyperbolae per infinitas hyperbolas
geometrice exhibit (Guido Grandi, 1703)
- Cesaro Summation: Ernesto Cesaro (1890)
- https://en.wikipedia.org/wiki/Grandi%27s_series