Infinite sum of logarithms

Question

Is there any closed form for this expression

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

Answer 1

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

Answer 2

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

Answer 3

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.

Answer 4

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.

Answer 5

The regularized value of this sum is

$\frac{1}{2} \ln \left(\frac{2 \pi }{\Gamma (x)^2}\right)$