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My question was inspired by this stackexchange question. For the last 90 minutes I have been trying to prove this formula due to Gregorio Fontana:

$$H_n = \gamma + \log n + {1 \over 2n} - \sum_{k=2}^\infty { (k-1)! C_k \over n(n+1)\ldots(n+k-1)}, \qquad \textrm{ for } n=1,2,3,\ldots,$$

where $H_n = \sum_{k=1}^n 1/k$ and the coefficients $C_k$ are the Gregory coefficients given by $${ z \over \log(1-z)} = \sum_{n=0}^\infty C_k z^k \qquad \textrm{ for } |z|<1.$$

It's a bit frustrating as it's something I recall proving as a student many years ago. I have a vague recollection that I began with something like:

$$H_n = \int_0^1 {1-(1-x)^n \over x } \textrm{d}x,$$

but my attempts to follow on from there have failed. Can you help?

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1 Answer 1

up vote 14 down vote accepted

Let's rewrite the formula, using $C_1=1/2$ as $$S_n=\gamma+\log n-H_{n-1}$$ where $$S_n=\sum_{k=1}^\infty\frac{(k-1)!C_k}{n(n+1)\cdots(n+k-1)}.$$ Use the formula $$\frac{(k-1)!}{n(n+1)\cdots(n+k-1)}=\int_0^1 x^{n-1}(1-x)^{k-1}dx$$ which is a special case of the beta integral. Therefore $$S_n=\int_0^1x^{n-1}\sum_{k=1}^\infty C_k(1-x)^k dx=\int_0^1\left(\frac{1}{1-x}+\frac{1}{\log x}\right)x^{n-1}dx.$$ When $n=1$ we get $$S_1=\int_0^1\left(\frac{1}{1-x}+\frac{1}{\log x}\right)dx=\gamma$$ by a well-known formula which can be derived from the aymptotic $$\int_t^\infty\frac{e^{-x}}{x}dx=-\log t-\gamma+O(t)$$ as $t\to0$, for the exponential integral.

By induction it suffices to consider the difference $$S_n-S_{n+1}=\int_0^1\left(x^{n-1}+\frac{x^{n-1}-x^n}{\log x}\right)dx =\frac1n-\int_0^\infty\frac{e^{-ny}-e^{-(n+1)y}}{y}dy.$$ Using the identity $$\int_0^\infty\frac{e^{-ay}-e^{-by}}{y}dy=\log\frac ba$$ for $0 < a < b$ which follows by integrating $e^{-xy}$ over the region $[a,b]\times[0,\infty)$ we get $$S_n-S_{n+1}=\frac1n-\log\frac{n+1}{n}.$$ The desired formula for $S_n$ now follows by induction.

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Excellent! Thanks for that. I spent a further 1hr 20mins on it this afternoon, without success. (My bin is full of scraps of paper covered in beta integrals!) Your method is different from the one I'm trying to recover from my distant past, but at least I have a proof now. Thanks again! –  Derek Jennings Sep 22 '10 at 17:42

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