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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.2012-08-19
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    How do you interpret the term $m=0$ ?2012-08-19
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    @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.2012-08-19
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    @Raskolnikov thanks i suppose m should be atleast 1 or more ofcourse m belonging to the set of natural numbers2012-08-19
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    The $m=0$ term is $\log(1) - 0 = 0$.2012-08-19
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    @RobertIsrael I am not sure i got why m = 0 we would get log(1) in that would n't that be meaningless if the series indexeing variable's upperbound is less than it's lower bound. If you could shed some more light on this i would be very greatfull.2012-08-19
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    Conventionally, the product of the empty set is $1$, just as the sum of the empty set is $0$.2012-08-19

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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.