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If possible, I would like to find a closed-form expression for $$ f(x)=\sum_{k=1}^\infty \frac{1}{x+k} $$ The series is so simple (possibly deceptively so) that I'm sure it has been studied before somewhere, but I don't know what to call it and where to find it.

What is this series called so I can learn more? I know I can expand it into a double series by expanding $\frac{1}{1-x/k}$ for $|x|<k$ and $\frac{1}{1-k/x}$ for $|x|>k$ and working on the partial series, which I did before in the special case of $|x|<1$, and the result yields a sum over Bernoulli numbers that I'm not familiar with.

Is a nice closed-form solution for this series known?

BGreen
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    The series diverges for any value of $x$. It is comparable to the harmonic series. – User8128 Dec 04 '20 at 02:05
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    Well, I should have seen that coming... thank you! I'll make a note to self not to ask questions when it is late. – BGreen Dec 04 '20 at 02:07
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    The symmetric form, that is $\lim_{N\to\infty}\sum_{k=-N}^{N} \frac{1}{x+k}$ is well-defined for $x\notin\mathbb{Z}$. See here, for instance. – Integrand Dec 04 '20 at 02:17

4 Answers4

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This series is also basically a part of the Hurwitz zeta function in the so-called ``s=1'' case. It was introduced in the later half of the 1800's and comes up in analytic number theory. There is no closed form and the theory is not elementary.

AndreyF
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Consider $$f_n(x)=\sum_{k=1}^n \frac{1}{x+k}=\psi(n+x+1)-\psi (x+1)=H_{n+x}-H_x$$ where appear the digamma function and the genarlized harmonic numbers.

Use the asymptotics $$H_p=\gamma +\log (p)+\frac{1}{2 p}-\frac{1}{12 p^2}+O\left(\frac{1}{p^4}\right)$$ and make $p=n+x$. Then continuing with Taylor series $$f_n(x)=\log(n)-\psi (x+1)+\frac{2x+1}{2n}-\frac{6 x^2+6 x+1}{12 n^2}+O\left(\frac{1}{n^3}\right)$$

Claude Leibovici
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$$ \frac 1 {\lfloor x \rfloor + k+1} \le \frac 1 {x+k} \le \frac 1 {\lfloor x \rfloor + k} $$ The two series whose terms are the first and the third expressions above are just tail ends of the harmonic series that diverges to $+\infty.$

Michael Hardy
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Using the following definitions $$ \psi(z) = -\gamma + \sum_{n=0}^\infty \left( \frac 1 {n+1} - \frac 1 {n+z} \right) $$ $$ \gamma = \lim_{n\,\to\,\infty} \left( -\ln n + \sum_{k=1}^n \frac 1 k \right) $$

one can conclude that \begin{align} & \sum_{k=1}^\infty \frac{1}{x+k}=\sum_{k=1}^\infty \frac{1}{k+1}-\gamma-\psi(x)+\frac{x-1}{x} \\[6pt] = {} & \lim_{n \to \infty} \ln(n)-\psi(x)+\frac{3x-2}{2x}. \end{align}

Anonymous
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