Is there a closed-form expression or a very good approximation for
$$
\sum_{i=0}^n \binom{n}{i} \log (i+1) \,?
$$
If the summands alternate, then there is a very close approximation, yet it feels like the alternation is a crucial ingredient.
So I was wondering if one can do better than estimating the sum by $\log(n+1) 2^n$.
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The question you cite doesn't give a closed form, and (by the discussion) a closed form for the alternating sum is extremely unlikely. – vonbrand Mar 14 '14 at 10:07
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@vonbrand OK, I changed the wording. The approximation in the linked question is very precise, that would be OK too. – john_leo Mar 14 '14 at 10:23
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The leading asymptotics is not much better than your guess. We have $$\ln(k+1)=\int_0^1\frac{x^k-1}{\ln x}\,dx$$ which gives $$\sum_{k=0}^{n}\binom{n}{k}\ln(k+1)=\int_0^1\frac{(1+x)^n-2^n}{\ln x}\,dx=2^n\big(\ln(n+1)-I_n\big),$$ where \begin{align} I_n&=\int_0^1\left[x^n-\left(\frac{1+x}{2}\right)^n\right]\frac{dx}{\ln x}\\\color{gray}{\left[x=1-\frac{t}{n}\right]\qquad}&=\int_0^n\left[\left(1-\frac{t}{n}\right)^n-\left(1-\frac{t}{2n}\right)^n\right]\frac{dt}{n\ln\left(1-\frac{t}{n}\right)}\\\color{gray}{[\text{apply DCT}]}\qquad&\underset{n\to\infty}{\color{red}{\longrightarrow}}\int_0^\infty\frac{e^{-t/2}-e^{-t}}{t}\,dt=\ln 2. \end{align}
Detailed asymptotics of $I_n$ can be obtained as follows. Write \begin{align} I_n&=\lim_{\epsilon\to 0}\int\limits_0^{1-2\epsilon}\left[x^n-\left(\frac{1+x}{2}\right)^n\right]\frac{dx}{\ln x}=\lim_{\epsilon\to 0}\left[\int\limits_0^{1-2\epsilon}\frac{x^n\,dx}{\ln x}-\int\limits_{1/2}^{1-\epsilon}\frac{2x^n\,dx}{\ln(2x-1)}\right]\\&=\int\limits_0^{1/2}\frac{x^n\,dx}{\ln x}+\lim_{\epsilon\to 0}\int\limits_{1/2}^{1-\epsilon}x^n\left(\frac{1}{\ln x}-\frac{2}{\ln(2x-1)}\right)dx-\lim_{\epsilon\to 0}\int\limits_{1-2\epsilon}^{1-\epsilon}\frac{x^n\,dx}{\ln x}\\&=\int\limits_0^{1/2}\frac{x^n\,dx}{\ln x}+\int\limits_{1/2}^{1}x^n\left(\frac{1}{\ln x}-\frac{2}{\ln(2x-1)}\right)dx+\ln 2. \end{align} The first integral is $\mathcal{O}(2^{-n})$; we forget it. The second one, after substitution $x=e^{-t}$, is $$-\int_0^{\ln 2}e^{-nt}\varphi(t)\,dt,\qquad\varphi(t)=e^{-t}\left(\frac{1}{t}+\frac{2}{\ln(2e^{-t}-1)}\right).$$ Here, Watson's lemma is applicable; expanding $$\varphi(t)=\frac{1}{2}-\frac{1}{4}t+\frac{1}{6}t^2+\frac{1}{48}t^3+\frac{23}{360}t^4+\frac{31}{480}t^5+\frac{131}{1680}t^6+\ldots,$$ we get $\color{blue}{I_n\asymp\ln 2-\dfrac{1}{2n}+\dfrac{1}{4n^2}-\dfrac{1}{3n^3}-\dfrac{1}{8n^4}-\dfrac{23}{15n^5}-\dfrac{31}{4n^6}-\dfrac{393}{7n^7}}-\ldots$
metamorphy
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Just adding an observed error term: $\frac{\ln(f(n)-h(n))}{n}$ seems to approach some constant between $0.6$ and $0.7$. $f(n)$ is the actual function and $h(n)$ is the approximation. – Varun Vejalla Oct 19 '19 at 07:08
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That then means that $f(n) \approx h(n) + e^{nk}$, where $k$ is the constant between $0.6$ and $0.7$. It may have been $\ln(2)$, so $f(n) \approx h(n) +2^n = 2^n(\ln(n+1)-\ln(2)+1)$. – Varun Vejalla Oct 19 '19 at 07:14
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1@automaticallyGenerated: $\frac{\ln(f(n)-h(n))}{n}=\ln 2+\epsilon_n$ implies only $f(n)=h(n)+2^n e^{n\epsilon_n}$. The last term is not implied to be $\asymp 2^n$. In fact I'm getting $f(n)\asymp 2^n\big(\ln\frac{n+1}{2}\color{blue}{+\frac{1}{2n}}-\frac{1}{4n^2}+\frac{1}{3n^3}\pm\ldots\big)$. – metamorphy Oct 19 '19 at 08:01
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That makes sense. Just curious, but how did you get the extra error terms? – Varun Vejalla Oct 19 '19 at 15:56
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Wow that's neat - never knew about Watson's lemma. So $f(n) = 2^n \left(\ln(n+1)-\ln(2)-\sum_{k=0}^{\infty} \frac{g^{(k)}(0)}{n^{k+1}} \right) $with $g(t) = e^{-t}\left(\frac{1}{t}+\frac{2}{\ln(2e^{-t}-1)}\right)$. – Varun Vejalla Oct 20 '19 at 04:13
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1I made this code f(n) = sum(k=1,n,binomial(n,k)log(k+1)); N=50; s = vector(N);x = vector(N); for (n=1,N-1, {s[n]=log(abs(f(n)-2^n(log(n+1)-log(2)+1/(2n)-1/(4n^2)))); x[n]=n;}); plothraw(x,s); your asymptotic seems valid https://pari.math.u-bordeaux.fr/gp.html – reuns Oct 30 '19 at 18:56