Ramanujan summation

Remarks

If $f$ is a polynomial, only finitely many derivatives are nonzero and the series in the definition is a finite sum.

Example: for $f(x) = x$, one has $f(0) = 0$, $f'(0) = 1$ and all higher derivatives vanish. So the Ramanujan sum of $1 + 2 + 3 + \cdots$ is $-\frac{B_2}{2} = -\frac{1}{12}$. For $f(x) = 1$, the Ramanujan sum of $1 + 1 + 1 + \cdots$ is $-\frac12$.

Not a regular method: for $f(x) = e^{-x}$ the series $\sum_{n \ge 1} e^{-n}$ converges to $\frac{1}{e - 1}$. From $\frac{t}{e^t - 1} = 1 - \frac{t}{2} + \sum_{k \ge 1} \frac{B_{2k}}{(2k)!} t^{2k}$ at $t = 1$ one gets $\sum_{k \ge 1} \frac{B_{2k}}{(2k)!} = \frac{1}{e-1} - \frac12$. Since $f^{(2k-1)}(0) = -1$, the Ramanujan sum is $-\frac12 + \frac{1}{e-1} - \frac12 = \frac{1}{e-1} - 1$. It differs from the ordinary sum by $\int_0^\infty e^{-x} \, dx = 1$.

Not stable: for $g(x) = f(x + 1)$ with $f = 1$, the Ramanujan sum of $\sum g(n)$ is $-\frac12$, not $-\frac12 - f(1) = -\frac32$.