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The following limit came up in the middle of solving a problem. Let $b$, $h$, and $k \geq 2$ be constants. If it helps simplifying the problem in any way, assume $b=2h$ and $k=2$.

I want to evaluate

$$\underset{m \rightarrow +\infty}{\lim} \prod_{i=0}^m \left( \left(1-\frac{(1-2i)b}{4h}\frac{1}{m+1} \right)^{\left( \dfrac{4}{k^2}\dfrac{1}{m+1} - \dfrac{(1-2i)b}{hk^2}\dfrac{1}{(m+1)^2} \right)} \right).$$

Numerical simulations show that this limit is a positive number less than 1. I would like to know how to evaluate this limit of product of infinite number of terms that approach 1.

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How do you know that it's possible to evaluate this limit in closed form? –  Christopher A. Wong Nov 26 '12 at 9:25
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(1) For any limiting problem involving a product, take logarithms first, then evaluate the limit, then exponentiate that answer. (2) Here, after you take logs, use the fact that $\log(1+x)$ is between $x$ and $x-x^2/2$ for $0<x<1$ (this comes from the power series for $\log(1+x)$), and then try to get the squeeze theorem involved. (3) Change $-(1-2i)$ to $+(2i-1)$.... –  Greg Martin Nov 26 '12 at 9:28
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Here's the best I can get so far: if $C = (1 - 2i)b/(4h)$, and $L$ is the desired limit, then we can write $$ \log{L} = \frac{4}{k^2} \sum_m \frac{m+1 - C}{(m+1)^2} \log\left( \frac{m+1 - C}{m+1} \right)$$ –  Christopher A. Wong Nov 26 '12 at 9:45
    
Thanks. Aside from directly evaluating the limit, what kind of approach can I take to show the limit exists in the interval $(0,1)$? –  tatterdemalion Nov 27 '12 at 21:51
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@tatterdemalion, given that one can turn the product problem into a series problem (as I wrote out above), you could try using various series convergence tests to try to show the limit exists. I haven't tried it myself, though. –  Christopher A. Wong Nov 27 '12 at 23:11

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