What is the asymptotic behavior of A103213 in OEIS?

Question

It's probably not at all hard—but at least right now it's not obvious to me—how to determine the asymptotic behavior of

$\sum_{k=1}^n \binom{n}{k} \frac{1}{k}$

After plotting, a plausible asymptote seems to be (a constant times) $\exp(2n/3)$. But I am not confident of this. — S Huntsman, Mar 29 '11 at 20:10
We have $$\sum_{k=1}^n {n\choose k}{1\over k} \geq \sum_{k=1}^n {n\choose k}{1\over k+1} = {2^{n+1}-2-n \over n+1}$$ so $\exp(2/3)$ is too small. — , Mar 29 '11 at 20:26
@Moron: I see that now, didn't parse the a(n) properly. My apologies. — S Huntsman, Mar 29 '11 at 22:12

score 13 · Accepted Answer · answered Mar 29 '11 at 20:33

13

Notice that $f(n) = \sum_{k=1}^n {n \choose k} \frac{1}{k} = \int_0^1 \frac{(t+1)^n - 1}{t}\, dt = \int_1^2 \frac{s^n - 1}{s-1} \, ds = \sum_{j=1}^{n} \frac{2^{j} - 1}{j}$. I think the leading term should be $2^{n+1}/n$

answered Mar 29 '11 at 20:33

Robert Israel

448,999

Nice! – Mar 29 '11 at 20:37
Oh, this is elegant. I'd considered Stirling but got caught up on something. But I didn't anticipate this. – S Huntsman Mar 29 '11 at 21:14
2

... and in fact I get $\frac{2^n}{n} (2 + \frac{2}{n} + \frac{6}{n^2} + \frac{26}{n^3} + \ldots) =$ (formally) $2^n \sum_{k=1}^\infty (-1)^k \frac{polylog(-k,2)}{n^{k+1}}$ – Robert Israel Mar 29 '11 at 23:14

score 9 · Answer 2 · answered Mar 29 '11 at 21:06

9

Here's a heuristic argument using probability.

Multiplying the OP's sum by $n/2^n$ gives $${1\over 2^n}\sum_{k=1}^n {n\choose k}{n\over k}=E\left({1\over \bar X_n}\right)$$ where $\bar X_n$ is the sample average of $n$ independent Bernoulli random variables with mean $1/2$ and we ignore the outcome $\bar X_n=0$.

By the law of large numbers, $1/\bar X_n\to 2$ almost surely, making it plausible that the left hand side of the equation above is approximately equal to 2.

answered Mar 29 '11 at 21:06

I love this argument! – Raskolnikov Mar 29 '11 at 21:09
This is interesting, thanks. – S Huntsman Mar 29 '11 at 21:12
@Raskolnikov Thanks. It needs some work to make it rigorous though. – Mar 29 '11 at 21:12

score 5 · Answer 3 · edited Apr 13 '17 at 12:20

(see the related question)

Asymptotically, the sum behaves like the integral $$\sum_{k=1}^n \binom{n}{k} \frac{1}{k} = \int_1^n dk \, \frac{n!}{k! (n-k)! \,k}.$$ For large $n$, you can approximate the integral by expanding the integrand around its maximum (attained at $k=n/2$). We have (using Stirling) $$\log \frac{n!}{k! (n-k)!} \sim n \log n - k \log k -(n-k) \log (n-k)$$ with the maximum at $k=n/2$. The expansion around $k=n/2$ reads $$n \log n - k \log k -(n-k) \log (n-k) = - \frac{2}{n} (k- n/2)^2.$$ Thereby, we can approximate $$\sum_{k=1}^n \binom{n}{k} \frac{1}{k} \sim \frac{2}{n}\binom{n}{n/2} \int_1^n dk e^{ -2 (k- n/2)^2/n} \sim \frac{2}{n} \frac{2^n}{\sqrt{\pi n/2}} \sqrt{\frac{\pi n}{2}} =\frac{2^{n+1}}{n} ,$$ where we used the fact that $\binom{n}{n/2} \sim 2^n/\sqrt{\pi n/2}$.

Nice, my original thought was to use Stirling but I must have made a mistake en route. Thanks. — S Huntsman, Mar 29 '11 at 21:12
I believe this problem allows you to use the same technique. — Douglas Zare, Mar 29 '11 at 23:32

Shai Covo · Answer 4 · 2011-03-29T21:22:30.783

Here is an exact lower bound, which, as is readily seen, approximately equal to $2^{n+1}/n$. By Jensen's inequality, since $1/x$ is convex, $$ \frac{{\sum\nolimits_{k = 1}^n {{n \choose k}} }}{{\sum\nolimits_{k = 1}^n {{n \choose k}} k}} \leq \frac{{\sum\nolimits_{k = 1}^n {{n \choose k}\frac{1}{k}} }}{{\sum\nolimits_{k = 1}^n {{n \choose k}} }}. $$ From this it follows straightforwardly that $$ \frac{{(2^n - 1)^2 }}{{2^{n - 1} n}} \le \sum\limits_{k = 1}^n {{n \choose k}\frac{1}{k}} . $$

EDIT: Hence, $$ \sum\limits_{k = 1}^n {{n \choose k}\frac{1}{k}} \geq \frac{{2^{n + 1} }}{n} - \frac{4}{n} + \frac{1}{{2^{n - 1} n}}. $$

What is the asymptotic behavior of A103213 in OEIS?

4 Answers4