# nth factor of $\frac {1\cdot2\cdot3\cdots( m-1) } {( z+1) ( z+2) \cdots ( z+m-1) }m^{s}$

Author of this book i am reading claims that the nth factor of $$\frac {1\cdot 2\cdot 3\cdots( m-1) } {( z+1)( z+2) \cdots( z+m-1) }m^{s}$$ is $\frac {n} {z+n}( \frac {n+1} {n}) ^{s}$ shouldn't it be just $$\frac {n} {z+n}( n+1) ^{s} ?$$

Edit: The author claims the product would look like $$\prod _{n=1}^{n=m-1}\frac {n} {z+n}\left( \frac {n+1} {n}\right) ^{s}.$$ I think (most likely incorrectly) that it should be written as $$\prod _{n=1}^{n=m-1}\frac {n} {z+n}\left({n+1} \right) ^{s}$$

i think i can see my mistake so i am going to close the post.

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Please define $n^{th}$ factor. – Aryabhata Mar 5 '12 at 19:24
Well the expression could be rewritten as a product, nth factor would be the nth term of that product. – Comic Book Guy Mar 5 '12 at 19:24
Even that is not good enough :-) As there are probably multiple ways of writing it as a product (multiplication being commutative and all). I am guessing you mean when we write out the product using $\prod_{n=1}^{m-1} f(n)$, we are talking about the two different $f(n)$ you mention in the question and whether the two products are the same. – Aryabhata Mar 5 '12 at 19:27
sorry that was a typo in the denominator, i guess s is some constant – Comic Book Guy Mar 5 '12 at 19:27
@Hardy: When you ask a question kindly add details to the questions so that the readers understand what is being asked and why is it tagged linear-algebra? – user17762 Mar 5 '12 at 19:28

The author is right.

What your product gives us is

$$\dfrac {1.2.3\ldots \left( m-1\right) } {\left( z+1\right) \left( z+2\right) \ldots \left( z+m-1\right) }(m!)^{s}$$

Note that it is $(m!)^s$ and not $m^s$.

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Yes u r right the author is indeed right the correct product expression should be $\left( \prod _{n=1}^{n=m-1}\dfrac {n} {z+n}\right)\left( \dfrac {m+1} {m}\right) ^{s}$ Do u concur ? – Comic Book Guy Mar 5 '12 at 19:45
@Hardy: No, that is not right. Now instead of $m^s$ you have $(\frac{m+1}{m})^s$. – Aryabhata Mar 5 '12 at 19:48
yes that was blunder indeed. thanks i think i got the answer i was looking for before i embarrass myself more. I'll conclude this thread. Thanks very much for your help. – Comic Book Guy Mar 5 '12 at 19:50
@Hardy: You are welcome. Mistakes are the best way to learn! So don't be embarrassed. – Aryabhata Mar 5 '12 at 19:52