I was looking for a way to assign a value to a divergent series and invented a formula myself. Here it is:

Suppose we want to assign a value to $$\lim_{n\to\infty} f(n)$$ where $f$ is some polymial. We introduce two symbols: $\omega$ and $\Delta$ with the property that $\omega * \Delta = 1$. These symbols represent infinity and an infinitesimal.

The Jippe value $f_\omega$ (my name is Jippe) is then obtained by the following formula: $$f_\omega = \sum_{k=0}^{\infty} \frac{\Delta^k}{k+1} f(\omega) \mod \omega $$ Here $\mod \omega$ means replacing all $\omega$ and $\Delta$ in the resulting expression with $0$.

As an example we calculate the Jippe value of $$\sum_{n=1}^{\infty} n$$ For every finite sum we have $$f(N) = \sum_{n=1}^{N-1} n = \frac{N(N-1)}{2}$$ Now $$f(\omega) = \frac{\omega(\omega-1)}{2} = \frac{\omega^2}{2}-\frac{\omega}{2}$$ $$f(\omega) \mod \omega = \frac{0}{2}-\frac{0}{2} = 0$$ $$\frac{\Delta}{2} f(\omega) = \frac{\Delta\omega^2}{4}-\frac{\Delta\omega}{4} = \frac{\omega}{4}-\frac{1}{4} = -\frac{1}{4} \mod \omega $$ $$\frac{\Delta^2}{3} f(\omega) = \frac{\Delta^2\omega^2}{6}-\frac{\Delta^2\omega}{6} = \frac{1}{6}-\frac{\Delta}{6} = \frac{1}{6} \mod \omega $$ $$f_{\omega} = -\frac{1}{4} + \frac{1}{6} = -\frac{1}{12}$$ This value is the same as the value of the Riemann Zeta function at $s=-1$ that also corresponds to this series. I have tested to see if the Jippe value is the same as the Riemann Zeta function for its values at $s=0$, $s=-2$ and $s=-3$ and found that $$\Bigl(\sum_{n=1}^{\infty} 1\Bigr)_\omega = \zeta(0) = -\frac{1}{2}$$ $$\Bigl(\sum_{n=1}^{\infty} n^2\Bigr)_\omega = \zeta(-2) = 0$$ $$\Bigl(\sum_{n=1}^{\infty} n^3\Bigr)_\omega = \zeta(-3) = -\frac{1}{120}$$ Take note that it matters where you start your function. If we take $$f(N) = \sum_{n=1}^{N-1} n,\quad g(N) = \sum_{n=1}^{N} n$$ Then $f_\omega = -\frac{1}{12}$ and $g_\omega = \frac{5}{12}$, which are not equal. To get the values alligned with the Riemann Zeta function, we need $f(1)=0$.

I also have a different formula that gives the same values at least on these negative integers. Here it is: $$f_\omega = \sum_{k=0}^{\infty} (-1)^k \frac{\Delta^k}{k+1} f(\omega+1) \mod \omega $$

I want to know

  1. Is this "result" new or does it exist already in some form.
  2. I want to know if this method is rigorous and unambiguous, that is: You get the same result no matter how you twist and turn. I know that series can be rewritten and then the Jippe value will change, but the corresponding function of the finite sum then also changes. By setting $f(1)=0$ and using the finite sum upto $N-1$ I think this is a well defined function.
  3. The series for which I have tested this all have a polynomial as their finite sums. Is there a "nice" formula for $\sum_{n=1}^{N} n^{-s}$ with $s \in \Bbb{C}$ or some specific values other than the above? (like s=1, s=1/2 or s=-1/2)

(I originally left the below out of the post as I think the post is very long)

The Jippe value can be generalized for $x^s, s \in \Bbb{C}$ as $(x^s)_\omega = \frac{\Delta^s}{s+1} \omega^s = \frac{1}{s+1}$. The formula proposed is then $$f_\omega = \sum_{k \in \Bbb{C}} \frac{\Delta^k}{k+1} f(\omega) \mod \omega $$

Also if a general function $f$ has $\lim_{x\to\infty} f(x)-g(x) = 0$ for a function $g$, we could say they have the same Jippe value. For convergent series the Jippe value then equal to the value of the series as $g(x)$ is a constant. A lot of values can then be calculated. For example I have $(e^x)_\omega = e + 1$ using Taylor Series, or $(-\frac{x^2}{1+x})_\omega = \frac{1}{2}$ using $\lim_{x\to\infty} (-\frac{x^2}{1+x} - (1-x)) = 0$

  • 1
    $\begingroup$ Your Jippe value for a polynomial $f(x) = a + bx + cx^2 + \cdots$ is equivalent to $a + \frac{b}{2} + \frac{c}{3} +\cdots$. I don't really understand what exactly you wish to know. $\endgroup$
    – orlp
    Commented Jul 26, 2017 at 19:56
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    $\begingroup$ Your Jippe value for polynomial $f(x)$ is also the antiderivate evaluated at 1 ($F(1)$). $\endgroup$
    – orlp
    Commented Jul 26, 2017 at 20:41
  • $\begingroup$ @orlp See here? $\endgroup$ Commented Jul 26, 2017 at 20:43
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    $\begingroup$ @SimplyBeautifulArt Essentially that's the same as my answer :) $\endgroup$
    – orlp
    Commented Jul 26, 2017 at 20:46
  • $\begingroup$ The math goes above my head, but if you integrate the generalized harmonic numbers (which are the partial sums of the sums $\dfrac{1}{k^s}$) and evaluate at $1$ using Wolfram Alpha you find that the Jippe values involve a simple fraction of zeta values. $\endgroup$
    – orlp
    Commented Jul 26, 2017 at 21:26

1 Answer 1


Your Jippe value for a polynomial $f(x) = a + bx + cx^2 + \cdots$ is equivalent to $a + \frac{b}{2} + \frac{c}{3} +\cdots$. This is easy to see as there is only a single value at a time where $\Delta \omega$ cancels.

Thanks to Faulhaber's formulas we have:

$$\sum_{k=1}^{n-1} k^p = - n^p + \frac{1}{p+1}\sum_{j=0}^p\binom{p+1}{j}B_jn^{p+1-j}$$

Where $B_j$ are the Bernoulli numbers of the second kind. This is a polynomial in $n$, so your Jippe value for the polynomial of $\displaystyle \sum_{k=1}^{n-1} k^p$ is equivalent to:

$$J = -\frac{1}{p+1} + \frac{1}{p+1}\sum_{j=0}^p\binom{p+1}{j}\frac{B_j}{2+p-j} $$

Since we have the following identity:

$$B_m = 1 - \sum_{k=0}^{m-1}\binom{m}{k}\frac{B_k}{m-k+1}$$

If we apply that identity with $m = p+1$ we get:

$$J = -\frac{B_{p+1}}{p+1}$$

Which is exactly the same as the zeta function for whole negative integers.

  • $\begingroup$ Kudos for this! This is nice $\endgroup$ Commented Jul 29, 2017 at 13:41

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