Let there be given an real-valued function $f(n)$ , with $n\in\mathbb{N}$ , $a,b \in\mathbb{R}$: $$ f(n+1) = a f(n) + b $$ What then is the value of $f(\infty)$ ?
As a physicist by education, being brought up with with limits and calculus, I would first derive the following, for $m\in\mathbb{N}$: $$ f(n+m) = a^m f(n) + \frac{1-a^m}{1-a} b $$ Consequently, but only for $|a| < 1$ : $$ f(\infty) = \lim_{m\to\infty} f(n+m) = b/(1-a) $$ And if $|a|\ge 1$ then this limit and so $f(\infty)$ does not exist.

However, once upon a time on the internet, sci.math to be precise, I've seen an argument like this - the original is in ASCII: $$ f(n+1) = a f(n) + b \quad \Longrightarrow \quad f(\infty) = f(\infty+1) = a f(\infty) + b \quad \Longrightarrow \quad f(\infty) = b/(1-a) $$ Thus, $f(\infty)$ is defined even if the sequence $f(n)$ is divergent. I've reproduced it as it says.

Now the question is: is the latter a valid argument in mathematics, and is $f(\infty)$ indeed defined, even if the sequence $f(n)$ is divergent?

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    $\begingroup$ $f(\infty)$ is sloppy notation. $\lim \limits_{n \to \infty} f(n)$ would be much preferred. $\endgroup$ – MCT Mar 12 '15 at 20:19
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    $\begingroup$ If $n\in\mathbb{N}$, then $f(\infty)$ is just meaningless; the only mathematical object to discuss is $\lim_{n\rightarrow\infty}f(n)$, which, if it exists, must be equal to $b/(1-a)$ by the argument you gave. On the other hand, if you want to define real-valued $f(\alpha)$ for all cardinals $\alpha$ with the same equation, then certainly $f(\alpha)=b/(1-a)$ must hold whenever $\alpha$ is infinite, since $\alpha + 1 = \alpha$. $\endgroup$ – mjqxxxx Mar 12 '15 at 20:20
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    $\begingroup$ @Soke: I agree, but $f(\infty)$ may have two interpretations here, which is the clue of the question. $\endgroup$ – Han de Bruijn Mar 12 '15 at 20:21
  • $\begingroup$ I would think of it like this; in your notation $f(\infty)$ is merely a symbol like $x$ which (in this case) represents the solution to $x=ax+b$ where $a\neq 1$. Whether this has much to do with the limit of a sequence depends, as you say, on the value of $a$. $\endgroup$ – Dan Mar 12 '15 at 20:22
  • $\begingroup$ The opening line describes $f(x)$ as "an arbitrary (say real-valued) function", but then restricts $f(x)$ to satisfying a very strict condition which admits only one function $f(x)$ once (say) $f(0)$ is specified. I would omit the word "arbitrary"; it suffices to consider real-valued functions that satisfy such a condition. $\endgroup$ – hardmath Mar 12 '15 at 20:34

It's a formal procedure to find the fixed point. Even if $|a|>1$, $\frac{b}{1-a}$ is still a fixed point of $f$. The difference is that if $|a| < 1$, $f$ converges to $\frac{b}{1-a}$ regardless of what $f(1)$ is, but if $|a|\geq1$ you only stay at the fixed point if that's where you started, while other points get moved off to infinity.

In fact, you'll notice that, formally, $\infty$ is also a fixed point. What this is saying is that $|a| < 1$ means that other points get sent to $\frac{b}{1-a}$, but if $|a|\geq1$ then other points get sent to $\infty$.

In short, the formal calculation does get a result that's meaningful in some sense, but not according to the usual definitions of limits. I'd draw a parallel to some interesting results you get from manipulating divergent series.

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    $\begingroup$ I really like the computational meaning / constructive content of your answer. Therefore the accept and (+1). Indeed : $f(n)=b/(1-a) \; \Longrightarrow \; f(n+1) = b/(1-a)$ . Fixed point is the keyword! $\endgroup$ – Han de Bruijn Mar 13 '15 at 16:09
  • $\begingroup$ @HandeBruijn: Thanks! $\endgroup$ – Tom Mar 14 '15 at 20:44

Infinity isn't a value, so saying $f(\infty)$ and $f(\infty + 1)$ is meaningless mathematically.


$$\lim_{x \to \infty} f(x) = \lim_{x \to \infty} f(x+1) = L,$$

but only if $L$ exists. This formalizes the argument you made.

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    $\begingroup$ The key to formalizing the argument in that way is to assume the limit $\lim_{x\to \infty} f(x)$ exists, so the argument cannot be relied upon to establish a valid limit when none exists. At least in part that seemed to be the thrust of the Question's closing paragraph. $\endgroup$ – hardmath Mar 12 '15 at 20:36
  • $\begingroup$ Infinity totally can be a value. What infinity isn't is a natural number. $\endgroup$ – Rahul Mar 13 '15 at 4:09

Long story short, your argument assumes that the function is convergent, i.e., that $f(\infty)$ is both finite and constant. But what if $f(\infty)$ is either infinite, or a set of values, instead of a single fixed point; i.e., $\sin(\infty)$ would be an entire interval, namely $[-1,+1]$, as opposed to a single fixed value.


As other answers have pointed out, $f(\infty)$ is defined if and only if you define it. Furthermore, you can define it in such a way that your formal manipulation is in fact a valid argument. So to me the question is not so much "is this valid?" but "is this useful?" Specifically, I'd like a definition such that:

  1. Your formal argument applies and correctly computes the limit.

  2. The definition applies to many different sorts of functions, and gives interesting answers for all of them.

I think such a definition exists. Moreover, there is a general principle for making interesting definitions of this type. This principle is analytic continuation.

The idea is simple: Express the limit or sum you want to study as a sum of a power series at a particular point. When the power series converges, you recover the usual answer. When it diverges, you try to sum the power series at a different point, forming a complex analytic function, and continue that analytic function around the singularity to your chosen point.

For instance, to the sequence $f(n)$ we could associate the power series $\sum_{n=1}^\infty f(n) x^n$. Observe that when $f(n)$ has a limit $f(\infty)$, as $x$ approaches $1$, this sum is $f(\infty)/(1-x)+ o(1/(1-x))$. So let's set:

$$g(x) = \sum_{n=1}^\infty f(n) x^n$$

$$ h(x) = (1-x) g(x)$$

and define $f(\infty)= h(1)$.

Now when the sum does not converge near $1$, but still converges near $0$, we can analytically continue $g(x)$. For the function you wrote down, it will always converge in some disc around $0$. Moreover, we may calculate from the recurrence relation:

$$ g(x) = \sum_{n=1}^\infty f(n) x^n = f(0) + \sum_{n=1}^{\infty} f(n+1) x^n = f(0) + ax g(x) + b \frac{x}{1-x} $$

$$g(x) = \frac{f(0) + b \frac{x}{1-x} }{1-ax}$$

$$h(x) = \frac{ f(0) (1-x) + bx }{1-ax}$$

These formulas describe $g$ and $h$ in the disc around $0$. Because these functions are analytic, they must also describe the analytic continuation of $g$ and $h$ wherever they are defined. Hence it is mathematically valid to compute:

$$h(1) = \frac{ f(0)(1-1) + b}{1-a} = \frac{b}{1-a}$$

This technique can be used to justify many other strange identities by appropriately defining the terms, such as the well-known "formula":

$$1+ 2+ 3+ 4 + \dots = \frac{-1}{12}$$

But, crucially, the analytic continuation of these sorts of sums is also crucial in many sorts of mathematical investigation - most famously the study of the Riemann zeta function.

It is also used in physics under the names "regularization" and "renormalization", but I don't know much about that.

When you are working with a "silly" formal manipulation that seems to produce an interesting result, don't ask yourself "is this rigorous?" but instead "how can I make this rigorous with appropriate definitions?" This mode of investigation has led to important insights in the past, and will lead to many more in the future. Maybe someday all symbolic manipulations used in physics will be given firm rigorous footing in their own mathematical theories that will have applications far beyond their original purpose.

  • $\begingroup$ $1+ 2+ 3+ 4 + \dots = -1/12$ ? Sorry, I don't buy it. $\endgroup$ – Han de Bruijn Mar 13 '15 at 15:51
  • $\begingroup$ @Han, you certainly don't have to buy it right now. But analytic continuation is something often done in applications. Maybe Will should have chosen the geometric series as a more accessible example. Consider $\frac1{1-2}=-1$, and think about the equivalent series expression. $\endgroup$ – J. M. is a poor mathematician Jun 28 '16 at 1:23
  • $\begingroup$ @J.M.: That would give $1+2+4+8+16+\cdots 2^n \cdots = -1$ . But the series diverges, therefore (IMO) it's nonsense. $\endgroup$ – Han de Bruijn Jun 29 '16 at 9:57
  • $\begingroup$ @J.M.: Yet such material is important for me, as is exemplified in : The Casimir Effect . The bottom line in that reference is : To be honest, I find all of the above calculations quite slippery; it feels like walking on thin ice. Any help would be quite welcome. $\endgroup$ – Han de Bruijn Jun 29 '16 at 10:05
  • $\begingroup$ @Han, indeed it looks like nonsense at first glance. But that is the precise point of analytic continuation: you want to find another expression that is sensible when your original expression is (supposedly) not. $\endgroup$ – J. M. is a poor mathematician Jun 29 '16 at 12:35

You ask the question, is $f(\infty)$ defined? Well $f(x)$ usually means take the number in the codomain of $f$ associated with the value $x$ in the domain of $f$. Since you defined $f$ to have a domain of $\mathbb N$, and $\infty\not\in\mathbb N$, this doesn't make sense. There are however several adhoc ways we can fix this, and look at the bottom of this answer, for how to solve this problem properly.

First some of the adhoc ways of solving this:
You obviously mean something different than applying the function with $\infty$. One way to define this faulty notation $f(\infty)$, could be using limits as you did: $$f(\infty):=\lim_{n\to\infty}f(n)$$ This interpretation makes reasonable sense, and results in different values for different $a$ and $b$, for example $a\ge1$ results in $f(\infty)=\pm\infty$ depening on $b$, and $|a|<1$ results in $f(\infty)=b/(1-a)$ and so on, I don't think finding these limits are at the heart of the problem.

Another way of interpreting this is saying that $\infty+1=\infty$ and plopping $\infty$ in the original definition, while ignoring that it's not defined for $\infty$. Then you get: $$ \begin{align} f(\infty+1) &= a f(\infty) + b\\ f(\infty) &= a f(\infty) + b\\ f(\infty)-a f(\infty) &= b\\ f(\infty)\cdot(1-a) &= b\\ f(\infty) &= \frac b{1-a}\\ \end{align} $$ And under this interpretation you get $b/(1-a)$ regardless of the values of $a$ and $b$.

When we see this phenomenom of several answers depening on the interpretation, there is always a clear answer on what the problem is and how to solve it:

The problem is that you have not defined what you mean properly, and the solution is to define what you mean properly.


An expression is defined if and only if it has a definition. The question of whether something is defined or not isn't very interesting; you can make anything be defined simply by defining it. The interesting question is whether or not an expression has a good definition­—one that makes sense in context.

So, does $f(\infty)$ have a "good" definition? It depends.

We can generally define functions however we want. If we define a function on the real numbers, we haven't automatically defined the function for $\infty$, because $\infty$ isn't a real number. But if I wanted to, I could say something like this:

I hereby define $f(\infty)$, for all functions $f$ defined on the real numbers, as the limit of $f(x)$ as $x$ increases without bound.

Aha, now $f(\infty)$ is defined (as long as the limit exists). But is this a good definition? Maybe, maybe not. It depends on what you're doing. Here's an example of a case where I would not say that this is a good definition.

Suppose we define the function $g$ like so: "For all real numbers $x$, $g(x) = (x + 1) - x$." Now, given all of our definitions, it would be somewhat reasonable to expect that $g(\infty) = (\infty + 1) - \infty$. (The reason this is a "reasonable expectation" rather than a certain fact is that we didn't say $g(x)$ is always $(x + 1) - x$; we only said that that's the case when $x$ is a real number.)

The problem is, we haven't defined $(\infty + 1) - \infty$ yet. Furthermore, it's not clear what a good definition of $(\infty + 1) - \infty$ is. (The general consensus is that there is no good definition of it.) So this reasonable expectation turns out to be wrong.

Here's another problem. What if I say this: "Define $h(x) = 3$ for all real numbers $x$, but define $h(\infty) = 4$." On its own, this is a perfectly legal definition, but this definition and our earlier definition of $f(\infty)$ are contradictory. So it's not legal to use both definitions simultaneously.

Sometimes our definition of $f(\infty)$ is perfectly fine, though. If $f$ is a rational function, then $f(x)$ always has a limit as $x$ approaches $\infty$ (though this limit might also be $\infty$). Rational functions extended in this way turn out to be pretty well-behaved.


You can always define $f(\infty)$, and you can make it whatever you like, and it doesn't have to have any relationship at all to other values of $f$ if you don't want it to.

However, given a function $f$, we usually only extend it to have a value at $\infty$ if it is a continuous extension — that is, defining $f(\infty)$ to be the value of the limit $\lim_{x \to \infty} f(x)$.

Sometimes, we have other reasons to define things in a certain way.

The particular argument you cited is suspicious — it is a motivation as to why we might define $f(\infty)$ in a particular way in that particular case, but not a proof. Note, incidentally, that if $a>0$ (and $b$ finite), then both $f(\infty) = \infty$ and $f(\infty) = -\infty$ are potentially relevant solutions to the equation the argument is solving.

Incidentally, the whole argument takes $a$ and $b$ as parameters; $f$ itself is a function of $a$ and $b$. After deciding to set $f(\infty)$ to the nice analytic formula $b/(1-a)$ in the case of $|a| < 1$, it often makes sense to take the analytic continuation — this formula remains valid for every pair of complex numbers $a,b$ with $a \neq 1$, so for many purposes it is reasonable to continue using the formula.

This is actually an argument form that gets a fair amount of use, especially in theoretical physics (as I understand it).

You have some thing you want to work with that, if it even makes sense at all, ought to be analytic. However, there are special cases that you can work with, such as restricting variables to sufficiently small magnitudes.

If you can work out an analytic formula for those special cases, then if the general case makes sense, it must also be given by the same analytic formula. And you can often you can point at the formula and say "yes, the general case makes sense, because the general case is that".


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