The numerical relation of the sum of two divergence series

Question

For these two series:

$1 + 2 + 3 + 4 + 5 +...$

$2 + 4 + 6 + 8 + 10 +...$

For each of the two series, since these numbers progress with no end, and the sum increases, it certainly cannot be finite. By this fact it becomes infinite. Hence, this quantity is so large that it is greater than any finite quantity. Thus I think the sum could only be an "infinite number", which is greater than any finite or assignable quantity. Moreover, every term of the second series is twice as large as the corresponding term of the first series(1 to 2, and 2 to 4, and n to 2n), it is reasonable to believe that the sum of the second series is twice as large as the first series, i.e. we get two infinite numbers and one is twice as large as the other.

so is there something wrong here?

Answer 1

The first thing I want to emphasize is that as these are sums of infinitely many terms, the meaning of this "operation" has to be defined; it doesn't follow automatically from the definition of finite sums. It turns out there are many different ways to think about these sums.

In short: There are interpretations in which they are connected to the same infinity (this would usually be taught in the first year of Calculus), ways in which they are connected to different infinities (as it seems you're looking for), and ways in which they could be related to different finite values (as Henry alluded to in a comment).

I'll try to summarize a lot of these, but I can't promise all the details will be accessible to everyone.

The same $\infty$

The sums of the first few terms keep going up, and the most common convention is to call the sum $\infty$ when that happens.

Calculus Treatment

In calculus, the value of the series (another word for "sum") is more or less defined to be the "limit of the sequence of partial sums", which ties it very closely to the definition of finite sum. In this case, the sequence of partial sums is $\left(1,1+2,1+2+3,1+2+3+4,\ldots\right)=\left(1,3,6,10,\ldots\right)$. But in a usual calculus class, this idea is a little confusing, because "limit of the sequence" has a couple definitions in calculus, depending on the situation.

The limit of a sequence $\left(a_1,a_2,\ldots\right)$ is said to be a real number (not any kind of infinity) $L$ (roughly) if the tails of the sequence (things like $\left\{a_{17},a_{18},\ldots\right\}$) eventually stay close to $L$. For example, the limit of the sequence $(-2,-1.1,-1.01,-1.001,-1.0001,\ldots)$ is $-1$, and the limit of the sequence $(1+\frac{\sin 1}{1},1+\frac{\sin 2}{2},1+\frac{\sin 3}{3},\ldots)$ is $1$ (as can be seen in a plot of the values: ). However, the numbers in $\left(1,3,6,10,\ldots\right)$ keep going up, and definitely don't stay close to any number. The limit of a sequence is said to be "$\infty$" (roughly) if the tails of the sequence eventually stay above any real number. (There is a similar definition for a limit to be $-\infty$.)

$\left(1,3,6,10,\ldots\right)$ stays above $10$ after the $4^{\text{th}}$ entry, above $100$ after the $13^{\text{th}}$ entry, above $1000$ after the $44^{\text{th}}$ entry, etc. So by this definition, the limit of the partial sums "is $\infty$", and the same can be said for $2+4+6+8+\cdots$.

However, I think this may be somewhat unsatisfying since "the limit is $\infty$" in a Calculus class is just shorthand for this "tails eventually stay above any real number" for two reasons: 1. $\infty$ here isn't anything like a number, it's just part of a shorthand. 2. This doesn't address any way in which $2+4+6+8+\cdots$ could be more than $1+2+3+4+\cdots$; there are lots of ways for the sequence of partial sums to grow.

Extended Reals

To address the first of those concerns a bit, without straying too far from the treatment in Calculus, there is a standard way to extend the real number line by adding on $\infty$ and $-\infty$ to the real numbers. They have some of the properties of the real numbers and help unify the three definitions of limit discussed earlier.

With Calculus, there's a quick-and-dirty description for unifying the definitions of limit: Define $\tan\frac{\pi}2=\infty$ and $\tan\left(-\frac{\pi}{2}\right)=-\infty$ (we won't worry about tangent outside of $[-\pi/2,\pi/2]$), and then redefine all limits in terms of the finite ones by: $\lim_{n}a_{n}:=\tan\left(\lim_{n}\arctan\left(a_{n}\right)\right)$. Since the inside limit is of arctans of numbers, it's a limit of a sequence in $\left(-\frac{\pi}2,\frac{\pi}2\right)$ so if it exists it's finite. If the limit of the $a_n$ would have been finite, then the limit of the arctans is some number in $\left(-\frac{\pi}2,\frac{\pi}2\right)$ and $\tan$ converts it back correctly. But if the limit of the $a_n$ would have been $\pm\infty$ by calculus convention, then the limit of the arctans would be $\pm\frac{\pi}2$, which is converted to $\pm\infty$ correctly by the definition we made.

Depending on your background, you can get a lot more out of the Wikipedia page for the extended reals.

Differing Infinities

While this isn't a usual convention for infinite sums, there are a couple of ways to look at the sequences of partial sums to say that the growth of $(2,2+4,2+4+6,\ldots)$ is somehow twice that of $(1,1+2,1+2+3,\ldots)$.

Asymptotics

One thing we can do to compare nonnegative sequences is to see what their ratio is, if it exists. If the limit is infinite, then in some sense the numerator function grows in a fundamentally faster way. If the limit is 0, the denominator function does. If the limit is finite, they grow at the same kind of rate (e.g. "quadratically"), and if the limit is 1, they grow at the same rate (the sequences are "asymptotic"). (There is a lot more to this sort of asymptotic analysis, especially when the limit doesn't exist, as seen in a table for big-O and related notations.)

In our case, the question is: what is the limit of the sequence $\left(\dfrac{2}{1},\dfrac{2+4}{1+2},\cdots\right)$? That sequence is actually the boring constant sequence $(2,2,2,\ldots)$, so the $(2,2+4,2+4+6,\ldots)$ is "asymptotically" (in the limit) twice as much as $(1,1+2,1+2+3,\ldots)$. This not revolutionary since the second sequence is exactly twice the first, but these ideas has much more general applications.

Unfortunately, functions are very much unlike numbers, so "one sum is an infinity that's twice the other one" is still out of reach.

Hyperreals

Just like there is the system of real numbers, there are extensions of reals (which contain some infinite numbers) known as systems (or fields) of hyperreal numbers. There is something called the "standard ultrapower construction of the hyperreals" which allows you to think about (equivalence classes of) sequences of real numbers as "hyperreal" numbers in their own right. (Wikipedia has one accessible account at https://en.wikipedia.org/wiki/Hyperreal_number#An_intuitive_approach_to_the_ultrapower_construction .)

Basically, the hyperreal number corresponding to $(1,1+2,1+2+3,\ldots)$ is an infinite number because the limit of the sequence (in the sense of Calculus) is $\infty$ (and similarly for $(2,2+4,2+4+6,\ldots)$). But in the hyperreals, these sequences represent numbers that you can do arithmetic with. The hyperreal number represented by $(2,2+4,2+4+6,\ldots)$ is twice that of $(1,1+2,1+2+3,\ldots)$, because $2=2*1$ and $2+4=2*(1+2)$ and $\ldots$. In fact, this only needs to hold for "most" of the entries, where the meaning of most is somewhat subtle.

One introduction to the hyperreals that I like, which uses a voting analogy and connects things to the discussion of asymptotics, is Terry Tao's Ultrafilters, Nonstandard Analysis, and Epsilon Management. If you want to use things like the infinite numbers that arise in the hyperreals without worrying about the ultrapower construction (or any other construction), then you can read Keisler's Elementary Calculus: An Infinitesimal Approach, which is intended to be at the level of an introductory calculus textbook.

Differing Finite Numbers?

There are senses in which it is natural to assign the values of $-\frac{1}{12}$ to the infinite sum $1+2+3+\cdots$ and $-\frac{1}{6}$ to the infinite sum $2+4+6+\cdots$ (not really thinking about the sequence of partial sums in either case). But where these weird values come from will take a bit to explain.

Regularization by Analytic Continuation

There is a nice formula for a sum a geometric series. When $|r|<1$ (even if $r$ is complex), $1+r+r^2+\cdots=\frac{1}{1-r}$ (in the sense of the limit of the sequence of partial sums). Certainly $1+2+4+\cdots=\infty$ in the calculus sense. But $\frac{1}{1-r}$ is a pretty nice function on the complex numbers (it's "meromorphic"), so we could declare by fiat that the formula applies for all $r$ except $r=1$. Then we have a sense in which $1+2+4+\cdots=\frac{1}{1-2}=-1$.

In general, if we have a sum that calculus says is $\infty$, then we might be able to find a collection of related sums that calculus says have regular complex values, and then find a nice function on the complex plane that agrees with those values while extending the domain as far as we can go (this is called analytic continuation). Then we declare by fiat that when the complex function makes sense, we should think of the sum as having that value.

Zeta Regularization

There is a general technique for finding nice complex functions for this purpose called Zeta regularization. A sum related to "$a_1+a_2+\cdots$" is "$a_1^s+a_2^s+\cdots$" which often converges to a complex number when the real part of $s$ is very negative. Then we can extend that function to many values of $s$ with analytic continuation, often getting a value at $s=1$, which we could call the value of the sum. Famously, when $a_1+a_2+\cdots$ is $1+2+\cdots$, the nice function is the Riemman zeta function, and we get (after some complex analysis work) answers like $1+2+\cdots=-\frac1{12}$, $1+1+1+\cdots=-\frac12$ and $1+4+9+\cdots=0$. This method leads to twice the zeta function when you start with $2+4+\cdots$ so that $1+2+\cdots=-\frac1{6}$ in this setting.

Exponential Regularization

Related to Zeta regularization is something called exponential regularization (for Dirichlet series), which we can use to get to the number $-\frac1{12}$ explicitly with only real calculus tools. Instead of looking at $1+2+\cdots$, let's look at $f(x)=e^{-x}+2e^{-2x}+3e^{-3x}+\cdots$. This converges when $x>0$, but not at $x=0$. However, we can still get a picture of this function near x=0 by doing some work with power series. First, note that if we integrate $f(x)$ from $\infty$ to $s$ term by term we get $\int f(x)=-e^{-s}-e^{-2s}-e^{-3s}-\cdots=-\dfrac{e^{-s}}{1-e^{-s}}$ (since it's a geometric series). Therefore, the original sum should be what we get when we differentiate this: $f(x)=\left.\dfrac{\mathrm d}{\mathrm d s}\,-\dfrac{e^{-s}}{1-e^{-s}}\right|_{s=x}=\dfrac{e^x}{\left(e^x-1\right)^2}$.

Now that we have a nice formula for $f(x)$, we can do some legwork to find some terms of a series for it (technically a Laurent series). One method would be to just use direct brute force with, for instance, the formula for the product of power series). However, with some intuition, we can open up another option (not really any nicer without a computer algebra system) by multiplying by $x^2$. $$\lim_{x\to0^{+}}\frac{x^{2}e^{x}}{\left(e^{x}-1\right)^{2}}=_{\text{L'H}}\lim_{x\to0^{+}}\frac{2xe^{x}+x^{2}e^{x}}{2e^{x}\left(e^{x}-1\right)}=\lim_{x\to0^{+}}\frac{2x+x^{2}}{2\left(e^{x}-1\right)}=_{\text{L'H}}\lim_{x\to0^{+}}\frac{2+2x}{2e^{x}}=1$$ Therefore, after multiplication by $x^2$, there's no longer a problem at $x=0$, and we can calculate the first few terms of a the Maclaurin series using Taylor's formula (taking the necessary limit each time), but the derivatives get complicated immediately.

Whichever method is used, we find that $f(x)=\dfrac{1}{x^2}+\dfrac{0}{x}-\dfrac{1}{12}+0x+\dfrac{x^2}{240}+\cdots$. We can't plug in $x=0$ because of the $\dfrac{1}{x^2}$ part, but if we just ignore that (and in general, just ignore all terms with negative powers of $x$), then we see the constant term $-\dfrac{1}{12}$ right there.

For a bit more discussion of some ways of getting finite values out of divergent series (and some mathematica code), you can read http://blog.wolfram.com/2014/08/06/the-abcd-of-divergent-series/. There are many more ways to sum divergent series on Wikipedia.

Answer 2

I don't know if i am wrong, but couldn' we simply say

$$S_n = 1+2+...+n=\frac{n(n+1)}{2}$$ and $$P_n=2(1+2+3+...+n)=2\frac{n(n+1)}{2}$$.

If we look at the ratio $$q_n=\frac{S_n}{P_n}=\frac{1}{2}\frac{S_n}{S_n}=\frac{1}{2}$$

Hence, couldn't we conclude ? $$q = \lim_{n \to \infty}q_n=\frac{1}{2}$$

So it seems to be quite obvious that even for $n \to \infty$ one could say, that the second series is twice as large as the first series.

Answer 3

What I think is wrong with your reasoning is the perception of infinity as a definable quantity (the statement "one infinity is twice of another" assumes two different infinities to be defined quantitatively. Equivalent statement would be that the two infinities are identifiable as specific points in the cartesian co-ordinate space). That is leading to your confusion.

Here are a few statements that would help you put things in perspective -

1. The two series grow indefinitely and 'get lost in the clouds'. Mathematicians are usually lazy enough to avoid writing this in so many words. They devised this nice symbol $\infty$ for this purpose.
2. The only quantitative comparisons that hold meaning involve measurable quantities, which by implication, must be finite. Hence, the moment you start thinking about the word 'twice', there's no point 'worrying what's above the clouds'.
3. Note that though the two series are infinite, they are completely defined using set builder definition thus: $$ S_1=\{x_n|x_{n+1}=x_n+1, x_1=1, n \in \Bbb N\} $$ $$ S_2=\{x_n|x_{n+1}=x_n+2, x_1=2, n \in \Bbb N\} $$ This is useful to realize that though we cannot make meaningful statements about the sum of all elements in $S_1$ and $S_2$ as measurable quantities, we can draw conclusions about their behaviour based on the observed patterns of their rising partial sums till the $t$'th term $$ S_{1,t}=\{x_n|x_{n+1}=x_n+1, x_1=1, n \in \Bbb N, n \le t\} $$ $$ S_{2,t}=\{x_n|x_{n+1}=x_n+2, x_1=2, n \in \Bbb N, n \le t\} $$ It is easy to see that sums of all elements in $S_{1,t}$ and $S_{2,t}$ are finite as long as $t$ is finite. And by using induction we find that $$\sum S_{2,t} =2* \sum S_{1,t} , \forall t \in \Bbb N$$ which is a meaningful quantitative comparison between the two series that gives a sense of 'HOW these two series are headed in the clouds'.

Note that definite conclusion could be drawn about the behaviour of the two series even though they are infinite because their definition is fully capture-able given any one element of the series and a set of well-defined mathematical operations that could be used to generate other elements

The word 'twice', and the quantitative comparison it implies, belongs to the domain of partial sums (these are finite) alone, and NOT to the convenience called $\infty$