65

This is a recreational mathematics question that I thought up, and I can't see if the answer has been addressed either.

Take a positive, real number greater than 1, and multiply all its roots together. The square root, multiplied by the cube root, multiplied by the fourth root, multiplied by the fifth root, and on until infinity.

The reason it is a puzzle is that while multiplying a positive number by a number greater than 1 will always lead to a bigger number, the roots themselves are always getting smaller. My intuitive guess is that the product will go towards infinity, it could be that the product is finite.

I also apologize that as a simple recreational mathematics enthusiast, I perhaps did not phrase this question perfectly.

glowing-fish
  • 623
  • 21
    Perhaps pose the question as, does the product, $\Pi_{i=1}^\infty (n)^{1/i}$ converge for any real numbers $n > 1?$ – Merkh Jul 13 '16 at 16:43
  • 23
    The symbolic representation helps, but I thought the question was stated fine as is. – Brian Tung Jul 13 '16 at 16:45
  • This is easily shown to diverge. A more interesting question is to consider only those roots that exceed a certain bound (such as 2) or some bound depending on $n$ (such as $1+1/\ln n$). – marty cohen Jul 13 '16 at 16:53
  • @Merkh I suspect it would be more appropriate to write an answer which explains what convergence of an infinite series means and why it is important. – Robert Frost Jul 13 '16 at 16:54
  • I only suggested a way to write the question down... not an answer. Sure, I suppose it would be useful defining convergence of an infinite series though. – Merkh Jul 13 '16 at 17:27
  • Have you tried logarithms? – user541686 Jul 14 '16 at 20:04

3 Answers3

83

This is an interesting question, but thanks to the properties of exponentiation, it can be easily solved as follows: $$\prod_{k\ge1}\sqrt[k]{n}=\prod_{k\ge1}n^{1/k}=n^{\sum_{k\ge1}\frac{1}{k}},$$ and as the exponent diverges (Harmonic series) the whole expression diverges, as long as $n>1$.

EDIT: the first equality is definition. For the second equality we need to argue why one can pass to the limit.

In other words: for any fixed $N$ the equality $$\prod_{k=1}^N n^{1/k}=n^{\sum_{k=1}^N \frac{1}{k}},$$ is valid, but why does it hold that $$\lim_{N\rightarrow \infty}n^{\sum_{k=1}^N \frac{1}{k}}=n^{\lim_{N\rightarrow\infty}\sum_{k=1}^N \frac{1}{k}}\quad ?$$ In this particular case it is a consequence of the fact that the map $x\mapsto n^x$ diverges to $+\infty$ as $x\rightarrow +\infty$ and by abuse of notation we write $n^{+\infty}=+\infty$

If conversely the exponent would converge to a finite limit, as for example in the case mentioned in the comment of IanF1, then this would be a simple consequence of the fact that the map $x\mapsto n^x$ is continuous.

b00n heT
  • 17,044
  • 1
  • 39
  • 52
  • @glowing-fish: For more information about the harmonic series, you might want to look at this Wikipedia article. – André Nicolas Jul 13 '16 at 17:08
  • 41
    This also implies that multiplying a number by its square root, fourth root, eighth root etc tends a limit of the original number squared. – IanF1 Jul 13 '16 at 19:12
  • 13
    Since it's non-obvious that you can move the infinite product/sum across the exponentiation (especially to someone with the OP's level of experience), I would write the whole string of equalities with $\lim_{k\to\infty}$ and sums/products from $i=1$ to $k$. – R.. GitHub STOP HELPING ICE Jul 13 '16 at 22:39
  • @R.. I agree, although I'd say that for someone with OP's level of experience, it is non-obvious that it should be non-obvious to someone with OP's level of experience! (That is, OP might take it for granted that of course you can do that, without realizing that in general generalizing some rules for infinite series isn't always easy or possible.) – JiK Jul 14 '16 at 09:31
  • 2
    @JiK Which is exactly the reason why this answer should be either edited or down voted. I'm currently doing neither but I really, really hate these kind of shortcuts since this is the upside down mentality that students often have. "Why isn't $(a+b)^2=a^2+b^2$ (over reals)?" That is the wrong question. Nothing is true unless you can prove it is, not everything is true until you find a counterexample. – DRF Jul 14 '16 at 12:29
  • I hope it is okay now – b00n heT Jul 14 '16 at 12:47
  • 1
    @b00nheT Prefect. Thank you and +1. – DRF Jul 14 '16 at 13:48
  • Continuity sufficient, monotony dispensable? Might be, looking at: $\lim\limits_{n \to \infty} \sin{\frac{(-1)^n}{n}} $ - continuity is (obviously) necessary, monotony isn't. – Imago Jul 14 '16 at 18:46
  • I don't think it's because of continuity (as it isn't continuous at infinity), but monotonicity instead. – Vim Jul 15 '16 at 10:29
20

What you are describing can be written as $$\prod_{n=1}^\infty{x^{1/n}}=x^1\times x^{1/2}\times x^{1/3} \times x^{1/4}...$$ Because to multiply a number to several powers is equivalent to adding their powers, this can be rewritten as $$x^{y}$$ where $$y={\sum_{n=1}^\infty{1/n}}$$

If we only added up the first $m$ terms of this, we get what's called the $m$th harmonic number. The harmonic numbers go 1, 3/2, 11/6, 25/12, ... Each one is about as big as the natural logarithm of $m$.

However since we're adding infinitely we are heading towards the "last" or "highest" harmonic number. Sadly, there is no greatest harmonic number and there is no point where they get ever-closer to some finite number so we are in some sense heading towards $x^\infty$.

We say that a series like this, i.e. which does not converge on some finite number "does not converge". And this makes it very difficult to quantify its size in a meaningful way.

However if however you want to get an idea of how quickly it diverges, you might imagine that the product of the first $m$ numbers is approximately equal to:

$$x^{\log_e{m}}$$

So if you had some number $p$ that measured the number of "infinite" terms you are multiplying together, then you might imagine your number is in some very limited sense $$x^{\log{p}}$$

From this you can observe that if you were to allow $x=e$ then your series equals $p$, so in a sense it reaches infinity at the same rate as you add terms. For $x=2$ it lags $m$ in approaching infinity, and for $x\geq3$ it is ahead of $m$ in approaching infinity.

Robert Frost
  • 9,620
  • 1
    I guess rewriting $x^{\log m}$ as $m^{\log x}$ gives a better idea of the polynomial growth rate of the series. – filipos Jul 13 '16 at 18:00
  • 16
    $\aleph_0$ and $+\infty$ are, for the most part, entirely unrelated. –  Jul 13 '16 at 18:59
  • @Hurkyl how are they related? – Robert Frost Jul 13 '16 at 20:03
  • 8
    "This can be rewritten as" is false unless you have some prior knowledge of convergence. – R.. GitHub STOP HELPING ICE Jul 13 '16 at 22:37
  • 3
    The bit about $\aleph_0$ strikes me as misleading at best. If I try to integrate $\int_{1}^{\infty}\frac{dx}x$, I definitely don't get $\log(|\mathbb R|)$ just because I integrated continuum many terms. Rather, the relevant notion of infinity is denoted by $\infty$ and is defined by adding a new maximal element to $\mathbb R$. The notion has nothing to do with how many elements are involved and the relevant topology more or less exactly expresses the intuition of capturing the behavior of, say, a function as its argument increases without bound (which is what's happening here) – Milo Brandt Jul 14 '16 at 02:56
  • @R.. that step only uses the rule that multiplication of like numbers to a power is equivalent to summation of their powers, which is entirely independent of convergence. I've deliberately avoided using convergence unexplained because of the OP's language. – Robert Frost Jul 14 '16 at 06:48
  • @MiloBrandt Ok I'll change the bit about $\aleph_0$, although $\sum_{x\in\mathbb{N}}{1_x}=\infty=\lvert\mathbb{N}\rvert$ – Robert Frost Jul 14 '16 at 08:49
  • 3
    @RobertFrost: No, because you have two expressions that don't even have values/meaning unless they converge. Hand-waving that away is at best confusing for non-experts who aren't going to know when it's okay and when it's not. – R.. GitHub STOP HELPING ICE Jul 14 '16 at 16:44
  • @R.. It's a known result since the 1700s that what works for every finite number works for the infinite. All I'm saying is $a^b\times a^c=a^{b+c} \forall a, b, c\in\mathbb{N}$. If you were to argue that the sum can't be evaluated because it diverges then you would have a point, but the sum of powers is definitely equivalent to the product. Independently of convergence we can nevertheless prove their identity since every characteristic each one has which is well-defined is identical to the other and neither has any characteristic which the other does not have. – Robert Frost Jul 14 '16 at 16:58
  • 3
    @RobertFrost: Your first sentence is wrong. I'm not going to read beyond that. – R.. GitHub STOP HELPING ICE Jul 14 '16 at 17:33
  • The only relation I'm aware of between $\infty$ and $\aleph_0$ is that when $\infty$ is used to denote an extended real number, it is is the supremum of the naturals in the extended reals, and when $\aleph_0$ is used to denote a cardinal number, it is the supremum of the naturals in the cardinal numbers. i.e. not directly related, but only indirectly in how they sit in relation to the natural numbers in their respective ordered classes. –  Jul 15 '16 at 08:08
  • Regarding one of the comments, note that when interpreting $\sum$ as a sum of extended real numbers in the only reasonable way, you still have $\sum_{x \in \mathbb{R}} 1 = \infty$, but when interpreted as a sum of cardinal numbers, you have $\sum_{x \in \mathbb{R}} 1 = |\mathbb{R}| \neq |\mathbb{N}|$. –  Jul 15 '16 at 08:10
  • @Hurkyl was it your intention to replace $\sum_{x\in\mathbb{N}}$ with $\sum_{x\in\mathbb{R}}$? Are you arguing $\sum_{x\in\mathbb{N}}1_x\neq\lvert\mathbb{N}\rvert$ or $\sum_{x\in\mathbb{N}}1_x\neq\infty$ ? – Robert Frost Jul 15 '16 at 08:32
  • In my previous comment, I am talking about sums where the real numbers are the index set. The difference in conclusions is meant to be contrasted with the assertion you had made about sums over the naturals; in particular that you once again equated $\infty$ with $|\mathbb{N}|$. –  Jul 15 '16 at 08:34
  • I am interpreting $|S|$ as the cardinality of the set $S$ -- if you meant to instead refer to the counting measure, then you do have $|\mathbb{N}| = |\mathbb{R}| = \infty$ and I withdraw my complaint about your equation involving $\sum_{x \in \mathbb{N}} 1$. –  Jul 15 '16 at 08:38
  • @Hurkyl I don't appreciate the difference to be honest but before I ask any more questions I think I'm hoping to understand how and why $\mathbb{R}$ was brought into it. – Robert Frost Jul 15 '16 at 09:01
  • @Hurkyl because the series is an ordered list and the number of elements is infinite and countable and is equal to $\lvert\mathbb{N}\rvert$ – Robert Frost Jul 15 '16 at 09:08
6

While the other answers are valid, they do tend to refer to previously established results, which (in a recreational context) is a bit frustrating. So here is a self-contained answer, for those who prefer that sort of thing.

First step: $$a^{1/2}\times a^{1/3}\times a^{1/4}\times…=a^{1/2+1/3+1/4+…}$$

Second step:$$\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+…\text{equals:}$$ $$\frac{1}{2}\ge\frac{1}{2}$$ $$…+\frac{1}{3}+\frac{1}{4}\gt\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$$ $$…+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\gt\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{1}{2}$$ $$\text{…and so on.}$$

That is, $\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+…$ is made up of infinitely many pieces, each of which is greater than $\frac12$. So it adds up to infinity, and so $$a^{1/2}\times a^{1/3}\times a^{1/4}\times…\text{ or}$$ $$a^{1/2+1/3+1/4+…}$$ is infinite if $a>1$, zero when $a<1$, and $1$ when $a=1$.

And not a $\sum$ sign anywhere to be seen!

Martin Kochanski
  • 3,862
  • 3
    We're dealing with an infinite series, however, and many identities that work with any finite series do not work with infinite series. So what proof do you have that your identity in step 1 works with an infinite series? – TLW Jul 14 '16 at 01:20
  • 4
    This is terribly non-rigorous. Each of your inequalities containing "$\cdots$" is literally "$\infty > \frac12=\frac12$". Doesn't say much, I'm afraid. One thing that is usually completely overlooked is the fact that an infinite series is not a sum at all, despite the suggestive notation. It is a limit of partial sums, which is a completely different animal. You can't "add up" infinitely many terms. – MPW Jul 14 '16 at 01:50
  • 1
    @MPW: It's true that can't get the result there only by repeatedly applying the binary addition operation to the terms, but it's wrong to say it's not a sum. (also, if we're being pedantic with your rationale, we would have to say things like $\sum_{n=a}^b n$ aren't sums either because they involve the recursion operator) –  Jul 15 '16 at 08:27
  • @Hurkyl: It's not a sum in the sense that it is not the result of adding up all of its terms. We refer to an infinite series as an "infinite sum" but this is a misnomer; the terminology is used because it is convenient and often such "infinite sums" behave like true sums. In reality, an "infinite sum" is a limit of a sequence of true sums: $S = \lim_{n\to\infty} S_n$. Each $S_n=\sum_{k=1}^na_k$ is a sum, but $\sum_{k=1}^{\infty}a_k$ is defined to mean $\lim_{n\to\infty} S_n$. $S$ is a limit of sums, not truly a sum itself, despite being called an "infinite sum". – MPW Jul 15 '16 at 13:00
  • @Hurkyl: Your last example is a perfectly well-defined sum, and it really is a sum of a finite number of terms (assuming $a$ and $b$ are integers). There is no recursion involved. It's just shorthand notation for adding up a specific finite list of real numbers (namely, the integers between $a$ and $b$, inclusive). – MPW Jul 15 '16 at 13:05
  • @MPW Although this is not rigorous, it seems like quite a trivial task to make it rigorous, no? – YiFan Tey Jun 24 '19 at 12:31