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In the analysis of an algorithm this statement has come up:$$\sum_{k = 1}^n\log(k) \in \Theta(n\log(n))$$ and I am having trouble justifying it. I wrote $$\sum_{k = 1}^n\log(k) = \log(n!), \ \ n\log(n) = \log(n^n)$$ and it is easy to see that $\log(n!) \in O(\log(n^n))$, but the "$\Theta$" part is still perplexing. I tried calculating the limit: $$\lim_{n \to\infty} \frac{\log(n!)}{\log(n^n)}$$but the only way that I thought of doing this was to substitute the Stirling approximation in the numerator, and it works, but this would only be justified if $$\lim_{n \to\infty}\frac{\log(n!)}{\log(\sigma(n))} = 1$$(with $\sigma(n) = \sqrt{2\pi} \cdot n^{\frac{1}{2} + n} \cdot e^{-n} $) and is it? It is certainly wrong that $$a_n \sim b_n \ \Rightarrow \ f(a_n) \in \Theta(f(b_n))$$ for a general (continuous) $f : \mathbb{R} \rightarrow \mathbb{R}$, since, for example, $a_n =n^2+n, b_n=n^2$ and $f(x) = e^x$ is a counterexample (since $\frac{n^2 + n}{n^2} \rightarrow 1$, but $ \frac{e^{n^2+n}}{e^{n^2}} \rightarrow +\infty$). So here are my questions:

  1. Is the statement true under certain hypothesis on $f$ (for example $f \in O(x)$ would be plausible), and in particular in the case of the $f(x) = \log(x)$?
  2. If not, how can I proceed in the initial proof, using Stirling, or otherwise?

I do not know (and don't want to use) anything andvanced on the Stirling approximation, such as error estimates; I know that $n! = \sigma(n)(1+ O(\frac{1}{n}))$, and not much more.

Any help is appreciated. Thank you.

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All you need to do is bound the sum from the left and right by Riemann sums, then take the corresponding Riemann integrals. – Qiaochu Yuan Jan 28 '12 at 19:45
Thank you, this solves 2. avoiding Stirling. If you can provide some insight into 1. I will happily accept your answer. – Emilio Ferrucci Jan 28 '12 at 19:56
Related question:… – Jonas Meyer Jan 28 '12 at 19:57
Actually, there is a more straight way of proving the approximation. Apply the trapezoid rule to get an approximation and then apply Euler-McLaurin's formula to bound that approximation. After taking appropriate limits, you might need to use the idea that the remainder $R$ in Euler-McLaurin's formula respect the relation $R_{n} = \lim_{n \to \infty} R_{n} + O \left( \frac{1}{n^m} \right)$, where $m$ is the power you are terminating. Taking limits and then exponentiating it, you will get the Stirling's formula! You don't need to find the error in approximation if you need $\Theta(n \log n)$ – Jalaj Jan 28 '12 at 20:13
@user135520 see here… – Emilio Ferrucci Nov 25 '15 at 13:43
up vote 11 down vote accepted

Here's another answer to question 2. By the Stolz–Cesàro theorem,


For a partial answer to question 1, here's a way to see that $a_n\sim b_n$ implies $\log(a_n)\sim\log(b_n)$ (under reasonable hypotheses such as $|\log(b_n)|>\varepsilon$ for some fixed $\varepsilon>0$ and sufficiently large $n$). Note that


This also implies that if $a_n\in\Theta(b_n)$ and $|\log(b_n)|\to\infty$, then $\log(a_n)\sim \log(b_n)$.

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+1 for elegance of the proof! – Jalaj Jan 28 '12 at 20:28
Very nice! I hadn't heard about the Stolz-Cesaro Theorem; but it makes a lot of sense thinking of it as the discrete analog of l'Hopital's rule. – Emilio Ferrucci Jan 28 '12 at 20:36

Okay, lets work out the maths.

By the trapezoid rule, $\ln n! \approx \int_1^n \ln(x)\,{\rm d}x = n \ln n - n + 1$. Now we find the error in the approximation, which one can compute by Euler–Maclaurin formula. After some crude arithmetic, you can compute that

$$ \ln (n!) - \frac{\ln n}{2}= n \ln n - n + 1 + \sum_{k=2}^{m} \frac{\mathcal{B}_k\,(-1)^k}{k(k-1)} \left( \frac{1}{n^{k-1}} - 1 \right) + S_{m,n}, $$ where $\mathcal{B}_k$ denotes the Bernoulli's number.

Now taking limits, you can compute $$\lim_{n \to \infty} \left( \ln n! - n \ln n + n - \frac{\ln n}{2} \right) = 1 - \sum_{k=2}^{m} \frac{\mathcal{B}_k(-1)^k}{k(k-1)} + \lim_{n \to \infty} S_{m,n}$$.

Now since $S_{m,n}$ satisfies the following property, $$S_{m,n} = \lim_{n \to \infty} S_{m,n} + O \left( \frac{1}{n^m} \right),$$ we get the approximation in the logarithmic form. Taking exponent on both the sides and plugging $m=1$, we get the result: $$n! = \sqrt{2 \pi n}~{\left( \frac{n}{e} \right)}^n \left( 1 + O \left( \frac{1}{n} \right) \right).$$

If you don't care about the $\sqrt{2 \pi n}$ term, I guess you can just use the approximation $\ln n! \approx \sum_{j=1}^n \ln j$ as follows:

$$\sum_{j=1}^n \ln j \approx \int_1^n \ln x \,{\rm d}x = n\ln n - n + 1 = \Theta(n \ln n)$$.

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Ah ok, this is a nice proof of Stirling's approximation. I don't know much about Bernoulli numbers; but this seems very interesting. However 2. of my question was already answered after the use of the trapezoidal rule, since $n \ln(n) - n +1 \in \Theta(n\ln(n))$ – Emilio Ferrucci Jan 28 '12 at 20:28
I know, but I thought I should pin down my thought process in the comment to your question more concretely. Otherwise, it sounded like a hand-waving argument and "obvious" and "easy to see" are two words that has no place in mathematics. – Jalaj Jan 28 '12 at 20:31
Yes, thanks; I was about to ask you to expand on the comment but you beat me to it by answering! I haven't quite figured out the details yet, but I will look at this more closely when I'm done with my upcoming exam. – Emilio Ferrucci Jan 28 '12 at 20:40

Put $s_n:=\sum_{k=1}^n\ln k$. Then $s_n\leq n\ln n$ and $$s_n=\sum_{k=1}^n\log (n-k+1)=n\log n+\sum_{k=1}^n\log\left(1-\frac{k-1}n\right)$$ and using $\log(1+t)\geq t-t^2/2$ we get
\begin{align*} s_n&\geq n\log n-\sum_{j=0}^{n-1}\frac jn+\frac{j^2}{2n^2}\\ &=n\log n-\frac{n-1}2-\frac{(n-1)(2n-1)}{6n}\\ &=n\log n-\frac{n-1}{2n}\left(n+\frac{2n-1}3\right)\\ &=n\log n-\frac{n-1}2\frac{5n-1}{3n}\\ &=n\log n-\frac{n-1}6\left(5-\frac 1n\right), \end{align*} so $$\frac{s_n}{n\log n}\geq 1-\left(1-\frac 1n\right)\frac 1{6\log n}\left(5-\frac 1n\right),$$ which is below bounded by a positive constant, for example $\frac 12$ for $n$ large enough.

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Thank you, this answer is very helpful. If you can provide any suggestions on how to answer 1., I will accept it. – Emilio Ferrucci Jan 28 '12 at 20:03
Except if a misunderstand something, we have $e^{\frac 1{n^2}}\sim e^{\frac 1n}$ but not $\frac 1{n^2}\sim\frac 1n$. – Davide Giraudo Jan 28 '12 at 20:08
Should it not be $s_n \geq n \log n - \sum_{j=0}^{n-1} \frac{j}{n} + \frac{j^2}{2n^2}$ because $t$ in your case is $\frac{-j}{n}$? From there you can argue that the term in the negative $\leq \Theta(n)$ and hence $s_n = \Theta(n)$. – Jalaj Jan 28 '12 at 20:08
@DavideGiraudo What I meant in the counterexample was: $\frac{n^2+n}{n^2} \rightarrow 1$, but $\frac{e^{n^2+n}}{e^{n^2}} = \frac{e^{n^2} \cdot e^n}{e^{n^2}} = e^n \rightarrow \infty$ – Emilio Ferrucci Jan 28 '12 at 20:12

An answer to question 1. involves slowly/regularly varying functions, whose study is sometimes summarized as Karamata's theory. Slowly varying functions are positive functions $L$ such that, for every positive $\lambda$, $L(\lambda x)/L(x)\to1$ when $x\to+\infty$.

To see that such functions make the implication in question 1. true, assume that $f$ is slowly varying and furthermore, to simplify things, that $f$ is nondecreasing. Then, if $a_n\sim b_n$ and $a_n,b_n\to+\infty$, $\frac12a_n\leqslant b_n\leqslant 2a_n$ for every $n$ large enough, hence $$ \frac{f(\frac12a_n)}{f(a_n)}\leqslant \frac{f(b_n)}{f(a_n)}\leqslant \frac{f(2a_n)}{f(a_n)}. $$ Since the lower bound and the upper bound both converge to $1$, this proves that $f(a_n)\sim f(b_n)$.

Thanks to Karamata's characterization theorem, the same kind of argument may be applied to the wider class of regularly varying functions. These are positive functions $L$ such that, for every positive $\lambda$, $L(\lambda x)/L(x)$ has a finite positive limit when $x\to+\infty$.

Hence the implication: $a_n\sim b_n$ implies $f(a_n)\sim f(b_n)$ if $a_n,b_n\to+\infty$, holds for every regularly varying function $f$.

Examples of slowly varying functions are the powers of $\log(x)$. Examples of regularly varying functions are the powers of $x$ (possibly, times a slowly varying function). Examples of non regularly varying functions are the exponentials of powers of $x$ (possibly, times a regularly varying function).

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Thank you, this is very interesting; I hadn't heard of these definitions. Actually regularly varying functions are more what I was after, since I was only asking that $f(a_n) \in \Theta(f(b_n))$ (not $f(a_n) \sim f(b_n)$). So polynomials do work (as one would expect), since they are regularly varying. – Emilio Ferrucci Feb 6 '12 at 12:42

All $n$ terms are smaller than or equal to $\log(n)$ and at least half are greater than or equal to $\log\left(\frac{n}{2}\right)$. So the sum is between $\frac{n}{2} \log\left(\frac{n}{2}\right)$ and $n\log(n)$. This gives the answer immediately.

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I seem to remember having seen this trick on the site not long ago... – Did Jan 14 '13 at 15:29
@did I just found the same argument at… , thanks. It is a completely standard method in undergrad algorithms 101 classes as well. – user54551 Jan 14 '13 at 18:26

Ok I came up with a (partial) answer to 1. It actually involves proving something stronger: $$\lim_{n \to \infty}(\log(b_n) - \log(a_n)) = \lim_{n \to \infty} \log \left(\frac{a_n}{b_n} \right) = \log \left(\lim_{n \to \infty} \frac{a_n}{b_n} \right) = \log(1) = 0$$ and when two sequences $x_n, y_n$, with $y_n \rightarrow +\infty$ (but probably definitively "far" from $0$ would have been enough) have the property that $y_n - x_n \rightarrow 0 \Rightarrow |y_n - x_n| \rightarrow0$, then: $$0 \leftarrow | y_n| \left|1 - \frac{x_n}{y_n} \right| \ \Rightarrow \ \frac{x_n}{y_n} \rightarrow 1$$ because if $$\exists \ \epsilon >0, \ \ \ \forall \ n_{\epsilon}, \ \ \exists \ n>n_{\epsilon}:\ \ \left|\frac{x_n}{y_n} - 1 \right| > \epsilon$$ then clearly $|y_n | \left|1 - \frac{x_n}{y_n} \right|$ could not converge to zero, since $|y_n| \rightarrow \infty$.
I think what made this proof possible is either the concavity of $\log$ or the fact that $\log \in O(x)$, but I get stuck when dealing with a general function $f$ with one of these properties. Anyway, the question is very much answered and I think Jonas' answer definitely deserves to be accepted :-)

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