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I was working on a summation and it was simplified to $$ \sum_{i = 1}^{n}H_{k - i} - \sum_{i = n + 1}^{i = 2n - 1} H_{k - i}\quad\mbox{where}\quad H_{n} = \sum_{k = 1}^{n}{1 \over k}\quad\mbox{where}\quad k = 2n $$

  • I tried using ${\tt Mathematica}$ to simplify this but unable to.
  • I also tried manually simplifying this by I can't come to a decent upper bound on this.
  • Does someone have any hints ?.
  1. Edit1: The above equation simplifies to $\log\left(\left(2n\right)!/\left(n!\, n!\right)\right)$. I am unable to understand $\qquad\,$ how to tightly bound this above and below.
  2. Edit2: @Jack pointed out that I made a mistake in simplification.
YuiTo Cheng
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am_rf24
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1 Answers1

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$$\sum_{i=1}^{n}H_{2n-i}-\sum_{i=n+1}^{2n-1}H_{2n-i} = \sum_{k=n}^{2n-1} H_k - \sum_{k=1}^{n-1} H_k=\sum_{k=1}^{2n-1}H_k-2\sum_{k=1}^{n-1}H_k \tag{1}$$ can be easily computed from $$\begin{eqnarray*} \sum_{k=1}^{N} H_k \stackrel{\text{SBP}}{=} N H_N - \sum_{k=1}^{N-1}\frac{k}{k+1}&=&NH_N-N+1+\sum_{k=1}^{N-1}\frac{1}{k+1}\\&=&(N+1)H_N-N\\&=&(N+1)H_{N+1}-(N+1).\end{eqnarray*} \tag{2}$$ In particular the RHS of $(1)$ equals

$$ 2n H_{2n-1}-(2n-1)- 2\left[n H_{n-1}-(n-1)\right]=\color{red}{2n(H_{2n}-H_n)}.\tag{3} $$ The RHS of $(3)$ behaves like $2n\log(2)-\frac{1}{2}+\frac{1}{8n}$ as $n\to +\infty$, while by Stirling's approximation $$ \log\frac{(2n)!}{n!^2}\sim 2n\log(2)-\frac{1}{2}\log(n)-\log\sqrt{\pi}-\frac{1}{8n}\tag{4} $$ as $n\to +\infty$, so $(1)$ and $(4)$ are different things.

Fun fact: the idea to approximate $$ \log\binom{2n}{n}=\sum_{k=1}^{2n}\log(k)-2\sum_{k=1}^{n}\log(k) $$ with $\sum_{k=1}^{2n}H_k-2\sum_{k=1}^{n}H_k=2n(H_{2n}-H_n)$ is more or less the key idea (together with creative telescoping) of my proof of Stirling's approximation.

Jack D'Aurizio
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  • What do you mean by (1) and (4) being different? – am_rf24 Jul 22 '20 at 15:47
  • @am_rf24: that the LHS of $(1)$ and the LHS of $(4)$ are different things. The part "the above equation simplifies to" of your question is not true. $(2)-2n\log 2$ is bounded, $(4)-2n\log 2$ is not bounded. – Jack D'Aurizio Jul 22 '20 at 15:53
  • You're correct. I made a mistake – am_rf24 Jul 22 '20 at 15:56
  • $$ \log\binom{2n}{n}=\sum_{k=n+1}^{2n}\log(k) - \sum_{k=1}^{n}\log(k) $$ is close, but not exactly equal, to $$\sum_{k=n+1}^{2n}H_k - \sum_{k=1}^{n}H_k,$$ since $$ H_n = \log(n)+\gamma+\color{red}{\frac{1}{2n}}+O(n^{-2}).$$ – Jack D'Aurizio Jul 22 '20 at 15:58
  • Got it @Jack. Thanks for pointing that out. So, that would be tightly bounded by O(n) right ? – am_rf24 Jul 22 '20 at 16:17
  • @am_rf24: yes, the harmonic sum is $n\log(4)+o(n)$, while the logarithm of the binomial coefficient is $n\log(4)+O(\log n)$. – Jack D'Aurizio Jul 22 '20 at 16:24