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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

$$\sum_{n=1}^{\infty}\frac{1}{P(n)}=-\sum_{i=1}^k\frac{H(-r_i)}{Q_i}$$

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

$$H(x)=\int_0^1\frac{1-t^x}{1-t}dt.$$

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

$$\sum_{n=1}^{\infty}\frac{1}{(x+\frac{1}{2})(x+\pi)(x+e)}=0.0841731$$

while

$$-\frac{H(1/2)}{(1/2-\pi)(1/2-e)}-\frac{H(-\pi)}{(\pi-1/2)(\pi-e)}-\frac{H(-e)}{(e-1/2)(e-\pi)}=0.0841731.$$

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.

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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, $$ \begin{align} \sum_{j=1}^\infty\frac1{P(j)} &=\sum_{j=1}^\infty\sum_{n=1}^k\frac{Q_n}{j-r_n}\\ &=\sum_{j=1}^\infty\sum_{n=1}^kQ_n\left(\frac1{j-r_n}-\frac1j\right)\\ &=\sum_{n=1}^kQ_n\sum_{j=1}^\infty\left(\frac1{j-r_n}-\frac1j\right)\\ &=-\sum_{n=1}^kQ_nH(-r_n) \end{align} $$

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

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