I want to prove Weierstrab's Infinite Product for the Gamma function. (without using theorems about infinite product or more difficult theorems)
By the Euler's limit formula for the Gamma function,
$\displaystyle \begin{eqnarray*} {1 \over \Gamma(x)} &=& \lim_{n \to \infty} { {x(x+1)(x+2) \cdots (x+n) } \over {n^x n!} } \\ &=& \lim_{n \to \infty} x n^{-x} \left( { {1+x} \over {1} } \right) \left( { {2+x} \over {2} } \right) \cdots \left( { {n+x} \over {n} } \right) \\ &=& \lim_{n \to \infty} x n^{-x} \prod_{k=1}^{n} \left( 1 + { x \over k } \right) \\ &=& x \lim_{n \to \infty} e^{-x \log{n} } \cdot e^{x \left( {1 \over 1} + {1 \over 2} + \cdots {1 \over n} \right) } \cdot e^{-x \left( {1 \over 1} + {1 \over 2} + \cdots {1 \over n} \right) } \cdot \prod_{k=1}^{n} \left( 1 + { x \over k } \right) \\ &=& x \lim_{n \to \infty} e^{ x \left( {1 \over 1} + {1 \over 2} + \cdots {1 \over n} - \log{n} \right) } \prod_{k=1}^{n} \left( 1 + { x \over k } \right) e^{- { x \over k} } \end{eqnarray*}$
I was stuck here not knowing whether or not: $\displaystyle \prod_{k=1}^{\infty} \left( 1 + {x \over k} \right) e^{- {x \over k} }$ is convergent.
But I found the following hint: $ \ln \left( 1 + {x \over n} \right) - {x \over n} \sim - {x^2 \over 2n^2}$ for large $n$
$\displaystyle \begin{eqnarray*} \lim_{n \to \infty} {{{x \over n} - \ln \left( 1 + {x \over n} \right)} \over {x^2 \over 2n^2}} &=& \lim_{n \to \infty} {{ {1 \over n} - {1 \over n} { 1 \over {1 + {x \over n}}} } \over {x \over n^2}} \\&=& \lim_{n \to \infty} {{ 1 - { 1 \over {1 + {x \over n}}} } \over {x \over n}} \\&=& \lim_{n \to \infty} n {{ {1 + {x \over n}} - 1 } \over x \left( {1 + {x \over n} } \right)} \\&=& \lim_{n \to \infty} {1 \over {1 + {x \over n} }} \\&=& 1 \end{eqnarray*}$
so $\displaystyle \exists \sum_{n=1}^{\infty}{x^2 \over 2n^2} \iff \exists \sum_{n=1}^{\infty} \left\{{x \over n} - \ln \left( 1 + {x \over n} \right) \right\} \iff \exists \prod_{k=1}^{\infty} \left( 1 + {x \over k} \right) e^{- {x \over k} } $
Am I correct?
If you are only interested in convergence but not evaluating the limit, then you don't have to mess with Gamma function at all. Just take log of this product and convert the partial product into a partial sum like $$\sum_{k=1}^N(\ln(1+\frac xk) - \frac xk)$$ Now can you use the hint you already know and kick start?