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Solve $\frac{1}{x} + \frac{1}{x+a} + \frac{1}{x+2a} + \cdots = b$ for $x$ given constants $a, b$.

I found that these types of series are called harmonic progressions and that there are no known formulas for partial sums, only approximations. However, the question I'm wondering is if this equation has complex solutions since it diverges for real values of $x$. If it diverges for complex values too, could someone give a proof?

Thanks

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  • $\begingroup$ The sum on the left will diverge regardless of $a$ and $b$. $\endgroup$ – Clayton Jun 13 '17 at 20:00
  • $\begingroup$ Wlog. $a$ is real and positive. If $|Na|\gg|x|$, we can estimate $\frac1{x+(N+k)a}\approx\frac1{ka}$ $\endgroup$ – Hagen von Eitzen Jun 13 '17 at 20:01
  • $\begingroup$ @HagenvonEitzen wouldn't that yield $$\frac1{x+(N+k)a}\approx\frac1{Na+ka}$$ ?? I would think we would need to know $|ka| \gg |Na| \gg |x|$ or the like to remove the $Na$ term $\endgroup$ – Brevan Ellefsen Jun 13 '17 at 20:03
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We may assume $a \ne 0$, and that none of the denominators $x + na$ is $0$. Then $$\dfrac{1}{x + n a} = \dfrac{1}{na} - \dfrac{x}{n a(x + n a)}$$ Now $\sum_n 1/(na)$ diverges. On the other hand, since $\frac{x}{na(x+na)} = O(1/n^2)$, $\sum_n x/(na(x+na))$ converges. Therefore $\sum_n \frac{1}{x+na}$ diverges.

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If $a$ is positive, comparing with the harmonic series here is no real (or complex) $x$ that makes the series $\sum_{n\in\Bbb{N}}\left(\dfrac{1}{x+na}\right)$ convergent. Same result is true when $a$ is negative as well. But if we restricted to finite sums, the equation $$\dfrac{1}{x} + \dfrac{1}{x+a} + \dfrac{1}{x+2a} + \cdots+\dfrac1{x+na} = b$$ can be solve as an algebraic equation of degree $n+1.$

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