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I've got a limit which puzzle me several days. The question is

$$ \lim_{n\to+\infty} \prod_{k=1}^n\left(1+\frac{k}{n^2}\right).$$

Can you help me? Thank you in advance

J. M. ain't a mathematician
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okBB
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    Surely if it has puzzled you several days, you have some work to show. – Emily Aug 16 '12 at 03:10
  • I've try to have a log transform first, and then reform it to calculus, but failed. My english is poor, I have difficults to show you in details. – okBB Aug 16 '12 at 03:30
  • @JackManey it is a very nice pedagogical discussion. I think it (the link) should put somewhere on the homepage. – vesszabo Aug 16 '12 at 14:46

2 Answers2

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Intuitively, we have

$$\log\left( 1 + \frac{k}{n^2} \right) = \frac{k}{n^2} + O\left(\frac{1}{n^2}\right) \quad \Longrightarrow \quad \log \prod_{k=1}^{n} \left( 1 + \frac{k}{n^2} \right) = \frac{1}{2} + O\left(\frac{1}{n}\right)$$

and therefore the log-limit is $\frac{1}{2}$.

Here is a more elementary approach: Let $P_n$ denote the sequence inside the limit. Then just note that

$$ P_n^2 = \left[ \prod_{k=1}^{n} \left( 1 + \frac{k}{n^2} \right) \right]^2 = \prod_{k=1}^{n} \left( 1 + \frac{k}{n^2} \right)\left( 1 + \frac{n-k}{n^2} \right) = \prod_{k=1}^{n} \left( 1 + \frac{1}{n}+\frac{k(n-k)}{n^4} \right). $$

Now fix $m$ and let $n \geq m$. Since $k (n-k) \leq \frac{1}{4}n^2$, we have

$$ \frac{k(n-k)}{n^4} \leq \frac{1}{4n^2} \leq \frac{1}{4mn}.$$

Thus we have

$$ \left( 1 + \frac{1}{n} \right)^n \leq P_n^2 \leq \left( 1 + \frac{1+(1/4m)}{n} \right)^n. $$

Thus taking $n \to \infty$,

$$e \leq \liminf_{n\to\infty} P_n^2 \leq \limsup_{n\to\infty} P_n^2 \leq e^{1+1/(4m)}.$$

Since $m$ is now arbitrary, we have $P_n^2 \to e$, or equivalently, $P_n \to \sqrt{e}$.

Sangchul Lee
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  • +1. A shortcut might be to use the fact that $(1+x_n/n)^n\to e^x$ for every $x_n\to x$, but the $1/(4m)$-trick is fine. – Did Aug 16 '12 at 10:03
  • @okBB the standard method to deal with product to take the logarithm of it (however in complex case you must be careful), so your idea was good. – vesszabo Aug 16 '12 at 15:37
  • I think you mean $\frac{k}{n^2} + O\left(\frac{1}{n^3}\right)$ in your first line, because otherwise the asymptotic isn't very interesting. – Najib Idrissi Aug 26 '12 at 12:11
  • @nik, I cannot understand why my asymptotic formula is uninteresting. Moreover, the error term cannot be $O(n^{-3})$ because it is actually $\Theta(k^2/n^4)$, which is bounded below by $O(n^{-2})$ for $k$ close to $n$. – Sangchul Lee Aug 26 '12 at 12:30
  • $k/n^2+O(k/n^2) = O(k/n^2)$, unless I'm mistaken. – Najib Idrissi Aug 27 '12 at 16:16
  • @nik, Still I can't understand why you mention $O(k/n^2)$. It's not what I used in the solution. – Sangchul Lee Aug 27 '12 at 20:59
  • That's my mistake, I meant to write $O(1/n^2)$. Sorry. What I'm saying is $k/n^2 + O(1/n^2) = O(1/n^2)$, when $n \to\infty$. – Najib Idrissi Aug 28 '12 at 09:11
  • @nik, your estimation would have been true if we were just considering it for a particular choice of $k$. Here, however, we are summing them for $k=1,\cdots,n$. My error term $O(1/n^2)$ is in fact a uniform bound depending only on $n$, but this bound does not apply to $k/n^2$, comsidering $k$ close to $n$. – Sangchul Lee Aug 28 '12 at 09:24
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As an alternative to @sos440's nice approach, note that $\mathrm e^{x-x^2}\leqslant1+x\leqslant\mathrm e^{x}$ for every $x$ in $[0,1]$. Hence the $n$th product $P_n$ is such that $S_n-T_n\leqslant\log(P_n)\leqslant S_n$, with $$ S_n=\sum_{k=1}^n\frac{k}{n^2}=\frac1n\sum_{k=1}^n\frac{k}{n},\qquad T_n=\sum_{k=1}^n\left(\frac{k}{n^2}\right)^2=\frac1{n^2}\sum_{k=1}^n\left(\frac{k}{n}\right)^2. $$ At this point, either one knows by heart the sum of the $n$ first integers and the sum of the $n$ first squares of integers, or one recognizes $S_n$ as a Riemann sum of the function $x\mapsto x$ on $[0,1]$, whose integral is $\frac12$, and $nT_n$ as a Riemann sum of the function $x\mapsto x^2$ on $[0,1]$. Either way, $S_n\to\frac12$ and $T_n\to0$, hence $\log P_n\to\frac12$ and $P_n\to\sqrt{\mathrm e}$.

Did
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