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I am wondering for $n > 1$ if the following series converges and if possible what it equates to.

$$\sum _{m=0}^{\infty }\dfrac {\log\left( \dfrac {\left( n+m-1\right) !} {\left( n-1\right) !}\right) -m} {\prod _{t=n}^{t=n+m-1}\log t}$$

I suspect it should sum to 1 as they are mutually exclusive probabilities, unless i made a mistake some where. I tried ratio test and wolfram alpha but did not have much success. Any help would be much appreciated.

Thanks in advance.

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For starters you can write your series as $$\sum\limits_{m = 0}^\infty {\frac{{\sum\limits_{k = n}^{n + m - 1} {\log k} - m}}{{\prod\limits_{k = n}^{n + m - 1} {\log k} }}} $$ and enjoy some "symmetry" of a kind. – Pedro Tamaroff Aug 19 '12 at 4:41
How do you interpret the term $m=0$ ? – Raskolnikov Aug 19 '12 at 4:46
@PeterTamaroff Thanks actually i knew that one, but was not sure what would have been the best version to present in the question. I suppose summation should precede factorial presentation. – Comic Book Guy Aug 19 '12 at 4:47
The $m=0$ term is $\log(1) - 0 = 0$. – Robert Israel Aug 19 '12 at 5:28
Conventionally, the product of the empty set is $1$, just as the sum of the empty set is $0$. – Robert Israel Aug 19 '12 at 18:23
up vote 6 down vote accepted

Well, it's certainly not going to always be $1$. Numerically, for $n=2$ I get approximately $1.52276115857537477668117216104$. In fact the partial sum up to $m=7$ is already greater than $1$.

The numerator of the $m$'th term is $O(m \log m)$, while the denominator is $\Omega(n^m)$ which grows much faster as $m \to \infty$ when $n > 1$. So the series converges.

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