# Infinite sum of logarithms

Is there any closed form for this expression

$$\sum_{n=0}^\infty\ln(n+x)$$

-
Is obviously $+\infty$ (when defined). – Martín-Blas Pérez Pinilla Jun 11 '14 at 10:05
log is defined over negative numbers, he never specified this was confined to the Reals. – hedgedandlevered Jun 11 '14 at 17:08

Write it as $\sum_{n=0}^k \ln(n+x)=\ln\left(\prod_{n=0}^k \left(n+x\right)\right)=\ln\left(\frac{\Gamma(k+x+1)}{\Gamma(x)}\right)$. If $k$ goes to $\infty$ the expression diverges, if $x>0$ or $x$ is not equal to a negative integer.

-

As others correctly mentioned, the expression diverges. Yet, if necessary, you can get quite good asymptotics: $$\sum_{k=1}^{n} \log (k+x) = \sum_{k=1}^{n} \log k + \sum_{k=1}^{n} \log (1+ \frac{x}{k}) \sim n \log n + \sum_{k=1}^{n} \frac{x}{k} = n \log n + x \log n \\ =(n+x) \log n$$

-

It's undefined whenever $x\leq 0$, and the series diverges to infinity whenever $x > 0$.

-

A simple way to see why it diverges for $x>0$ is as follows: $$\sum_{n=0}^\infty \ln(n+x) \ge \sum_{n=1}^\infty\ln(1+x) = \infty$$ as an infinite sum of a constant positive argument is always infinite (I have used monotonicity of the logarithm here).

For $x\le 0$, $\ln(0+x)$ is not defined, hence the entire sum is not defined.

-
How about this $$\sum_{n=1}^\infty \ln(n) = \frac{1}{2}\ln(2\pi)$$ if we add x > 0 to the inside logarithm? – Nguyen Jun 11 '14 at 15:01
I don't understand what's written there. The sum on the left diverges as well. – Bach Jun 11 '14 at 15:38
Hi Bach, I got that from here mathworld.wolfram.com/LogarithmicSeries.html. Is that correct? – Nguyen Jun 11 '14 at 16:36
@Nguyen: from the page you linked: Note that the first two of these are divergent in the classical sense, but converge when interpreted as zeta-regularized sums. – Andrea Corbellini Jun 11 '14 at 17:28