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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.

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    $\begingroup$ Perhaps pose the question as, does the product, $\Pi_{i=1}^\infty (n)^{1/i}$ converge for any real numbers $n > 1?$ $\endgroup$
    – Merkh
    Commented Jul 13, 2016 at 16:43
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    $\begingroup$ The symbolic representation helps, but I thought the question was stated fine as is. $\endgroup$
    – Brian Tung
    Commented Jul 13, 2016 at 16:45
  • $\begingroup$ 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 $1+1/\ln n$). $\endgroup$
    – marty cohen
    Commented Jul 13, 2016 at 16:53
  • $\begingroup$ @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. $\endgroup$
    – it's a hire car baby
    Commented Jul 13, 2016 at 16:54
  • $\begingroup$ 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. $\endgroup$
    – Merkh
    Commented Jul 13, 2016 at 17:27

3 Answers 3

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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.

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  • $\begingroup$ @glowing-fish: For more information about the harmonic series, you might want to look at this Wikipedia article. $\endgroup$
    – André Nicolas
    Commented Jul 13, 2016 at 17:08
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    $\begingroup$ This also implies that multiplying a number by its square root, fourth root, eighth root etc tends a limit of the original number squared. $\endgroup$
    – IanF1
    Commented Jul 13, 2016 at 19:12
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    $\begingroup$ 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 $k$. $\endgroup$
    – R.. GitHub STOP HELPING ICE
    Commented Jul 13, 2016 at 22:39
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    $\begingroup$ @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. $\endgroup$
    – DRF
    Commented Jul 14, 2016 at 12:29
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    $\begingroup$ @b00nheT Prefect. Thank you and +1. $\endgroup$
    – DRF
    Commented Jul 14, 2016 at 13:48
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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.

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    $\begingroup$ $\aleph_0$ and $+\infty$ are, for the most part, entirely unrelated. $\endgroup$
    – user14972
    Commented Jul 13, 2016 at 18:59
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    $\begingroup$ "This can be rewritten as" is false unless you have some prior knowledge of convergence. $\endgroup$
    – R.. GitHub STOP HELPING ICE
    Commented Jul 13, 2016 at 22:37
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    $\begingroup$ 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) $\endgroup$
    – Milo Brandt
    Commented Jul 14, 2016 at 2:56
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    $\begingroup$ @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. $\endgroup$
    – R.. GitHub STOP HELPING ICE
    Commented Jul 14, 2016 at 16:44
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    $\begingroup$ @RobertFrost: Your first sentence is wrong. I'm not going to read beyond that. $\endgroup$
    – R.. GitHub STOP HELPING ICE
    Commented Jul 14, 2016 at 17:33
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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!

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    $\begingroup$ 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? $\endgroup$
    – TLW
    Commented Jul 14, 2016 at 1:20
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    $\begingroup$ 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. $\endgroup$
    – MPW
    Commented Jul 14, 2016 at 1:50
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    $\begingroup$ @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) $\endgroup$
    – user14972
    Commented Jul 15, 2016 at 8:27
  • $\begingroup$ @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". $\endgroup$
    – MPW
    Commented Jul 15, 2016 at 13:00
  • $\begingroup$ @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). $\endgroup$
    – MPW
    Commented Jul 15, 2016 at 13:05

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