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I'm trying to solve the following limit: $$L=\lim_{n\to\infty}\frac{(n^2-1)(n^2-2)\cdots(n^2-n)}{(n^2+1)(n^2+3)\cdots(n^2+2n-1)}=\lim_{n\to\infty}\prod_{k=1}^n\frac{n^2-k}{n^2+2k-1}$$ Since originally, the limit yields an $1^\infty$ indeterminate expression, my first idea was taking logarithms: $$\log L = \lim_{n\to\infty}\log\left(\prod_{k=1}^n\frac{n^2-k}{n^2+2k-1}\right)=\\ =\lim_{n\to\infty}\sum_{k=1}^{n}\log\left(\frac{n^2-k}{n^2+2k-1}\right)=\\=\lim_{n\to\infty}\sum_{k=1}^n\log\left(\frac{1-\frac{k}{n^2}}{1+\frac{2k-1}{n^2}}\right)=\\ =\lim_{n\to\infty}\sum_{k=1}^n\left(\log\left(1-\frac{k}{n^2}\right)-\log\left(1+\frac{2k-1}{n^2}\right)\right)=\\=\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\left(\log\left(1-\frac{k}{n^2}\right)^{n^2}-\log\left(1+\frac{2k-1}{n^2}\right)^{n^2}\right)=\\ =\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\left(\log\left(1+\frac{1}{-\frac{n^2}{k}}\right)^{-\frac{n^2}{k}(-k)}-\log\left(1+\frac{1}{\frac{n^2}{2k-1}}\right)^{\frac{n^2}{2k-1}(2k-1)}\right)$$

Now if I could do the limit of the terms inside the sum I'd have: $$\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\left(\log\left(1+\frac{1}{-\frac{n^2}{k}}\right)^{-\frac{n^2}{k}(-k)}-\log\left(1+\frac{1}{\frac{n^2}{2k-1}}\right)^{\frac{n^2}{2k-1}(2k-1)}\right)=\\=\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\left(\log e^{-k}-\log e^{2k-1}\right)=\\ =\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\left(-k-2k+1\right)=\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\left(-3k+1\right)$$ and then the limit would be trivial. The question is: is that allowed? Why? If not, how could I proceed? Thanks!

Hercule Poirot
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  • Are you aware of Riemann sums? – DatBoi Nov 23 '20 at 16:13
  • @DatBoi Yes! But I haven't been able to find a suitable partition and function to apply them here. – Hercule Poirot Nov 23 '20 at 16:14
  • Reassessing the question, I dont think Riemann sum is helpful here. Your approach seems to be ok and you can refer this – DatBoi Nov 23 '20 at 16:29
  • @DatBoi It seems the post you are referring to states that I can move the limit inside any continuous function. The problem here is that the sum is not continuous and, above all, depends on the limit variable n. In general one cannot move the limit inside the sum, although perhaps in this case there is some condition I don't know about that allows it. – Hercule Poirot Nov 23 '20 at 16:37
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    I'm working on it. I'll let you know in a while – DatBoi Nov 23 '20 at 16:38

2 Answers2

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Let's take the first part

$$\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\log\left(1+\frac{1}{-\frac{n^2}{k}}\right)^{-\frac{n^2}{k}(-k)} = -\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n k\log\left(1+\frac{1}{-\frac{n^2}{k}}\right)^{-\frac{n^2}{k}}$$

as example. Note that $-\frac{n^2}{k}$ tends to $\infty$ for any $k=1,2,\cdots,n$, so we can use Taylor's expansion

$$\left(1+\frac{1}{x}\right)^x = e - \frac{e}{2x} + O\left(\frac{1}{x^2}\right) \quad (x \to \infty)$$

to obtain

$$\begin{aligned} \lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\log\left(1+\frac{1}{-\frac{n^2}{k}}\right)^{-\frac{n^2}{k}(-k)} &= -\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n k\log\left[e+\frac{ek}{2n^2}+O\left(\frac{k^2}{n^4}\right)\right] \\ &= -\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n \left[k + k\log\left(1+\frac{k}{2n^2}+O\left(\frac{k^2}{n^4}\right)\right)\right] \\ &= -\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n \left[k + k\left(\frac{k}{2n^2}+O\left(\frac{k^2}{n^4}\right)\right)\right] \\ &= -\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n k\left[1+O\left(\frac{1}{n}\right)\right], \end{aligned}$$

where we have used $\frac{k}{n^2}=O\left(\frac{1}{n}\right)$. So the term $O(1/n)$ can be discarded.

Note: An alternative solution is given here:

Lemma. Given $f(0) = 0$ and that finite $f'(0)$ exists, let

$$a_n = \sum_{k=1}^n f\left(\frac{k}{n^2}\right) = f\left(\frac{1}{n^2}\right) + f\left(\frac{2}{n^2}\right) + \cdots + f\left(\frac{n}{n^2}\right),$$

then

$$\lim_{n\to\infty} a_n = \frac{f'(0)}{2}.$$

Proof. First notice that

$$f'(0) = \lim_{x \to 0} \frac{f(x)-f(0)}{x-0} = \lim_{x \to 0} \frac{f(x)}{x},$$

and using the definition of limit, $\forall \varepsilon > 0$, $\exists \delta > 0$, s.t. $\forall 0 < x < \delta$,

$$f'(0) - \varepsilon < \frac{f(x)}{x} < f'(0) + \varepsilon.$$

Particularly for $x > 0$, we have $(f'(0)-\varepsilon)x < f(x) < (f'(0)+\varepsilon)x$.

Now pick $N \in \mathbb{N}$, s.t. $N > \dfrac{1}{\delta}$. For $n > N$,

$$\frac{k}{n^2} \leq \frac{1}{n} < \frac{1}{N} < \delta, \quad k=1,2,\cdots,n,$$

and hence

$$(f'(0) - \varepsilon) \cdot \frac{k}{n^2} < f\left(\frac{k}{n^2}\right) < (f'(0) + \varepsilon) \cdot \frac{k}{n^2}, \quad k=1,2,\cdots,n.$$

Taking summation over $k$ gives

$$\frac{f'(0)-\varepsilon}{2} \cdot \frac{n+1}{n} < \sum_{k=1}^n f\left(\frac{k}{n^2}\right) < \frac{f'(0)+\varepsilon}{2} \cdot \frac{n+1}{n}.$$

Since $\dfrac{n+1}{n} \to 1$ as $n \to \infty$, then $\exists N_1 \in \mathbb{N}$, s.t. $\forall n > N_1$,

$$\frac{f'(0)-\varepsilon}{2} - \frac{\varepsilon}{2} < \sum_{k=1}^n f\left(\frac{k}{n^2}\right) < \frac{f'(0)+\varepsilon}{2} + \frac{\varepsilon}{2},$$

that is, $\dfrac{1}{2}f'(0)-\varepsilon < \sum\limits_{k=1}^n f\left(\frac{k}{n^2}\right) < \dfrac{1}{2}f'(0)+\varepsilon$. Therefore we have

$$\lim_{n\to\infty} a_n = \lim_{n\to\infty} \sum\limits_{k=1}^n f\left(\frac{k}{n^2}\right) = \frac{f'(0)}{2}.$$

Return to your problem, we have

$$I = \lim_{n\to\infty}\prod_{k=1}^n\frac{n^2-k}{n^2+2k-1} = \lim_{n\to\infty}\prod_{k=1}^n\frac{n^2-k}{n^2+2k} = \lim_{n\to\infty}\prod_{k=1}^n \frac{1-\frac{k}{n^2}}{1+2\frac{k}{n^2}} = \lim_{n\to\infty} e^{\sum_{k=1}^n f(k/n^2)},$$

where $f(x) = \ln\left(\frac{1-x}{1+2x}\right)$ and the second equality comes from

$$1 \leftarrow \left(1-\frac{1}{n^2}\right)^n \leq \prod_{k=1}^n \frac{n^2+2k-1}{n^2+2k} = \prod_{k=1}^n \left(1-\frac{1}{n^2+2k}\right) \leq 1.$$

Therefore $I = e^{\frac{f'(0)}{2}} = \boxed{e^{-\frac{3}{2}}}$.

Mephestopheles
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  • Hi @Mephestopheles, thanks for your answer. I've been trying to understand it and there are two steps I can't seem to figure out. In the first method, how does the logarithm disappear in the second-to-last step? And in the second method, when you do the $I^{-1}$ trick, why does the second term inside the product vanish? – Hercule Poirot Nov 25 '20 at 13:28
  • The logarithm disappears owing to the Taylor's expansion for $\log(1+x)$. The second term inside $I^{-1}$ is the reciprocal of $(n^2-1)(n^2-2) \cdots (n^2-n)$ and naturally tends to $0$. – Mephestopheles Nov 26 '20 at 04:47
  • Aaah ok, now I get the first one. However, I still can't see the second one. I mean, once you expand the product it's true that every term with a $\frac{1}{n^2-k}$ factor will tend to zero, but as $n$ tends to infinity there will be infinitely many of them, so how can you be sure that the sum of all of those terms will still have limit zero? – Hercule Poirot Nov 26 '20 at 09:27
  • Okay, are you are wondering whether the equation $\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n f(k,n) = \lim_{n\to\infty}\frac{1}{n^2} \sum_{k=1}^n \left[ \lim_{n\to\infty} f(k,n) \right]$ holds? If the limit $\lim_{n\to\infty} f(k,n) = \tilde{f}(k)$ exists for any choice of $k$ (such as $f(k,n)=n^2\log(1+\frac{k}{n^2})$ in your case), then you can write $f(k,n) = \tilde{f}(k) + o(1)$. Now the summaztion becomes $\sum_{k=1}^n \tilde{f}(k) + o(n)$, well you know how to proceed. – Mephestopheles Nov 26 '20 at 10:40
  • @Hercule Poirot In the second method, it is an infinite product (rather than sum) of terms tending to zero, i.e., $0^{\infty}$, which is still $0$. Btw, when you are confusing about limit inside limit, Taylor's expansion as above may be helpful~ – Mephestopheles Nov 26 '20 at 10:49
  • Thanks, I'm beggining to understand. However, I'm still confused about the product. From your answer I think you are saying that $\lim_{n\to\infty}\prod_{k=1}^n \left( \frac{n^2+2k}{n^2-k}-\frac{1}{n^2-k} \right)=\lim_{n\to\infty}\prod_{k=1}^n \frac{n^2+2k}{n^2-k}-\lim_{n\to\infty}\prod_{k=1}^n\frac{1}{n^2-k}$ and the second product would be $0^\infty=0$. But I think that's not the case since $(a+b)(c+d)=ac+bc+ad+bd\neq ac+bd$. So there would be more terms aside from $\prod_{k=1}^n \frac{n^2+2k}{n^2-k}$ and $\lim_{n\to\infty}\prod_{k=1}^n\frac{1}{n^2-k}$, wouldn't there? – Hercule Poirot Nov 26 '20 at 20:35
  • Also, what you said in your comment about $\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n f(k,n) =\lim_{n\to\infty}\frac{1}{n^2}\left(\sum_{k=1}^n \tilde{f}(k) + o(n)\right)$ was really helpful! That and the Taylor expansion really made the matter clear to me. Thanks so much for your help! :) – Hercule Poirot Nov 26 '20 at 20:43
  • So sorry that I made such a silly mistake. I've edited my answer, and the determination of limit $\prod_{k=1}^n \frac{n^2+2k-1}{n^2+2k}$ still involves 'limit inside limit', i.e., $\lim_{n\to\infty} \left(1-\frac{1}{n^2}\right)^{n^2 \cdot \frac{1}{n}} = \lim_{n\to\infty} \frac{1}{e^{1/n}}$, but of a much simpler form. – Mephestopheles Nov 27 '20 at 10:41
  • Don't worry, thanks a lot! Now everything is clear. You've helped me very much. – Hercule Poirot Nov 27 '20 at 10:55
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$\require{cancel}$ $$L=\lim_{n\to\infty}\frac{\cancel{n^{2n}}(1-\frac1{n^2})(1-\frac2{n^2})\cdots(1-\frac{n}{n^2})}{\cancel{n^{2n}}(1+\frac1{n^2})(1+\frac3{n^2})\cdots(1+\frac2n-\frac1{n^2})}=\lim_{n\to\infty}\frac{e^{\frac{-1}{n^2}}e^{\frac{-2}{n^2}}...e^{\frac{-n}{n^2}}}{e^{\frac{1}{n^2}}e^{\frac{3}{n^2}}...e^{\frac{2n-1}{n^2}}}=\lim_{n\to\infty}\frac{e^{\frac{-1}{n^2}{(\sum_{r=1}^n r)}}}{e^{\frac{1}{n^2}{((2\sum_{r=1}^n r)-n)}}} \\=\lim_{n\to\infty}e^{\frac{-1}{n^2}((3\sum_{r=1}^n r)-n)}=\lim_{n\to\infty}e^{((-3\frac{1}{n}\sum_{r=1}^n \frac{r}{n})+\frac1n)}=\lim_{n\to\infty}e^{((-3\int_0^1xdx+\frac1n)}\to e^{\frac{-3}{2}}$$

DatBoi
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  • Thanks, @DatBoi, I find your answer really easy to follow and the result is the same I get with my method. However, my question remains, why can you do the limits of the bases in the first step while not doing the limits of the exponents? I mean, in order to get all those e you have already solved a limit, but only of a part of the expression. Why is that allowed? – Hercule Poirot Nov 23 '20 at 17:59
  • I dont quite understand what you mean by "to get all those e you have already done a limit, but only of a part of the expression" – DatBoi Nov 23 '20 at 18:02
  • I mean in the first step, after you have extracted the $n^{2n}$ factors and simplified, I believe you have used the fact that $\lim_{n\to \infty}\left(1+\frac{1}{-\frac{n^2}{k}}\right)^{-\frac{n^2}{k}}=e$. That is the same trick I used after I introduced the limit inside the sum. You did it with a product instead, but the fact remains that the product (or the sum in my case) still depends on the number n. Is the limit of the product the product of the limits, even when there are a variable number of factors? – Hercule Poirot Nov 23 '20 at 18:11
  • @HerculePoirot The number of factors is fixed- $n$. The reason why I chose to do this rather than simplifying it using logarithms(i.e. keeping it as a product instead of sum) is because it is easier to understand that limit of product is the product of limits and limit of quotient is the quotient of limits. – DatBoi Nov 23 '20 at 18:19
  • I'm sorry, @DatBoi, but I don't see it. If the number of factors is a fixed number $n$ then every factor is a constant! (Since they only depend on that very same number $n$). So I don't understand why we can do the limits that result in all those $e$. – Hercule Poirot Nov 23 '20 at 18:28
  • @HerculePoirot When I say that the number of factors is fixed and is equal to $n$ , I mean that there are countably infinite factors. – DatBoi Nov 23 '20 at 18:43