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Is there any closed form for this expression

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

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1  
Is obviously $+\infty$ (when defined). –  Martín-Blas Pérez Pinilla Jun 11 at 10:05
    
log is defined over negative numbers, he never specified this was confined to the Reals. –  hedgedandlevered Jun 11 at 17:08

4 Answers 4

up vote 5 down vote accepted

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.

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

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

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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.

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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 at 15:01
    
I don't understand what's written there. The sum on the left diverges as well. –  Bach Jun 11 at 15:38
    
Hi Bach, I got that from here mathworld.wolfram.com/LogarithmicSeries.html. Is that correct? –  Nguyen Jun 11 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 at 17:28

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