Background: This is from question 12.3.2 from Arfken, Weber, Harris Mathematical methods. The definition of the The Euler-Maclaurin integration formula in the text is: enter image description here

The formula is taken from n=1(instead of 0) to $\infty$ and rearranged to solve for the Euler-Mascheroni constant: $\sum_{n=1}^{\infty}\frac{1}{n}-\ln(n)$ and given to you in the question (where $f(x)=\frac{1}{n}$ above):enter image description here

Ignoring the final error term integral in 12.57, the lower terms for n=1 marked in red seem to be left off: $-\frac{f(1)}{2}$(added to balance $+\frac{f(1)}{2}$) and the $-\sum_{k=1}^{N}\frac{B_{2k}}{2k}$ Why?

  • $\begingroup$ Start the series at $n=1$ to avoid division by zero, then apply Euler-Maclaurin. $\endgroup$ Apr 29, 2017 at 14:13
  • 1
    $\begingroup$ @sim And after starting at n=1 why are $-1/2 f(1)$ and $-f^{2p-1}(1)$ not there? $\endgroup$ Apr 29, 2017 at 14:17

1 Answer 1


Knuth's article can be found here. Essentially the idea is to compare the approximation for finite $n$ with the equality obtained for $n$ infinite (at this point I want to take issue with the use of "$=$" in the book, which is obviously nonsense since $\gamma$ is constant and the right-hand side clearly has different values for differing values of $N$).

So, finite $n$: $$ \sum_{s=1}^{n} \frac{1}{s} = \log{n} + \frac{1}{2} + \frac{1}{2n} + \sum_{k=1}^N \frac{B_{2k}}{2k}\left( 1 - \frac{1}{n^{2k}} \right) - \int_1^n \frac{B_{2k+1}(\{x\})}{x^{2k+2}} \, dx \tag{1} $$ (Knuth has written the remainder in a format that makes it easier to deal with here, using the periodic versions of the Bernoulli polynomials. Since it's essentially an integration by parts and a sum, it really is equality. It's going to disappear anyway, so don't worry too much about the exact form.) Taking $n\to \infty$, (1) becomes $$ \gamma = \frac{1}{2} + \sum_{k=1}^N \frac{B_{2k}}{2k} - \int_1^{\infty} \frac{B_{2k+1}(\{x\})}{x^{2k+2}} \, dx, $$ and subtracting (1) from this gives $$ \gamma = \sum_{s=1}^{n} \frac{1}{s} - \log{n} - \frac{1}{2n} + \sum_{k=1}^N \frac{B_{2k}}{2k}\frac{1}{n^{2k}} - \int_n^{\infty} \frac{B_{2k+1}(\{x\})}{x^{2k+2}} \, dx. $$ The remainder integral is much smaller than the other terms if we take $n$ a reasonable size (Knuth uses Stirling's formula on an explicit expression for $B_{2k+1}(\{x\})$ to exhibit this), and thus one ends up with the expression in the book.

  • $\begingroup$ The motivation being to reduce the number of terms in the error integral it looks like (by adding explicit exact terms at the beginning $(\sum^n\frac{1}{s}$). $\endgroup$ Apr 29, 2017 at 15:20
  • $\begingroup$ In (1), why is the third plus sign not a minus sign? $\endgroup$ Apr 29, 2017 at 15:31
  • $\begingroup$ Odd derivatives of $1/x$ have minus signs. $\endgroup$
    – Chappers
    Apr 29, 2017 at 15:32
  • $\begingroup$ Oh, but $B_{2n+1}(0)=0$. These are the even B's. You are right in the final answer I just don't understand why yet. I may find the answer in the primary source. $\endgroup$ Apr 29, 2017 at 15:33
  • $\begingroup$ It's $B_{2k} f^{(2k-1)}$. Even Bernoulli number, odd derivative. $\endgroup$
    – Chappers
    Apr 29, 2017 at 15:35

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