This is question based on a pattern I have noticed while using mathematica. Let $P(x)$ be a polynomial with real, simple, negative roots $r_n$ ($n:1,2,...,k$) and define

$$Q_n=\lim_{x\to r_n}\frac{P(x)}{x-r_n}.$$

Then the pattern seems to be


where $H(x)$ denotes the analytic continuation of the harmonic nubmers:


For example, for $P(x)=(x+\frac{1}{2})(x+\pi)(x+e)$, we have




For clarification, the reason the roots must be negative is to sidestep any convergence issues in the infinite sum. My question is, does anyone know where I can find a proof of this fact or what some good topics/terms might be to search for? I have been handily using this fact in my current research, but upon starting the writeup today, I realized that I had never actually proved it.

If it helps, I believe this source deals with a more generalized version of this problem (as the digamma function and harmonic number are intimately linked). Unfortunately, they did not include a description of their coefficients (as far as I can tell) and only showed that it was a combination of a finite number of digamma functions.


1 Answer 1


Using the reciprocals of your $Q_n$, $$ \begin{align} Q_n &=\lim_{x\to r_n}\frac{x-r_n}{P(x)} \end{align} $$ The Heaviside Method for Partial Fractions says. $$ \frac1{P(x)}=\sum_{n=1}^k\frac{Q_n}{x-r_n} $$ Note that if $k\ge2$, $$ \begin{align} \sum_{n=1}^kQ_n &=\lim_{x\to\infty}\sum_{n=1}^k\frac{x\,Q_n}{x-r_n}\\ &=\lim_{x\to\infty}\frac{x}{P(x)}\\[6pt] &=0 \end{align} $$ Therefore, $$ \begin{align} \sum_{j=1}^\infty\frac1{P(j)} &=\sum_{j=1}^\infty\sum_{n=1}^k\frac{Q_n}{j-r_n}\\ &=\sum_{j=1}^\infty\sum_{n=1}^kQ_n\left(\frac1{j-r_n}-\frac1j\right)\\ &=\sum_{n=1}^kQ_n\sum_{j=1}^\infty\left(\frac1{j-r_n}-\frac1j\right)\\ &=-\sum_{n=1}^kQ_nH(-r_n) \end{align} $$

  • $\begingroup$ Perfect, this is exactly what I was looking for $\endgroup$
    – QC_QAOA
    Jan 11, 2019 at 17:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.