Why is $\prod_{n=1}^\infty \dfrac{(1+1/n)^s}{1+s/n}$ convergent for $s$ complex with $Re(s)>0$?

Question

Let $s$ be a complex number with $Re(s)>0$. I am trying to show that $$f(s) = \prod_{n=1}^\infty \frac{(1+1/n)^s}{1+s/n}$$ converges.

My approach was to look at the sum $$\sum |1-f(s)|$$ converges, but I got nowhere with this. Could anyone either give a hint or provide a solution to why this infinite product is convergent?

Edit I corrected a typo which made the infinite product divergent.

Answer 1

About the updated question: $$\begin{eqnarray*} \prod_{n=1}^{N}\frac{\left(1+\frac{1}{n}\right)^s}{1+\frac{s}{n}}&=&\frac{(N+1)^s N!}{\prod_{n=1}^{N}(n+s)}\\[0.2cm]&=& \frac{\Gamma(s+1)\Gamma(N+1)}{\Gamma(N+s+1)}(N+1)^s\\[0.2cm]&=&N(N+1)^s B(s+1,N)\end{eqnarray*}$$ can be written as $$\begin{eqnarray*} &&N(N+1)^s \int_{0}^{1}x^{s}(1-x)^{N+1}\,dx\stackrel{x\mapsto\frac{z}{N+1}}{=}\frac{N}{N+1}\int_{0}^{N+1} z^s\left(1-\frac{z}{N+1}\right)^{N+1}\,dz\\[0.2cm]&=&\frac{N}{N+1}\int_{0}^{+\infty}z^s\cdot \underbrace{\left(1-\frac{z}{N+1}\right)^{N+1}\mathbb{1}_{(0,N+1)}(z)}_{\to e^{-z}\text{ pointwise}}\,dz\end{eqnarray*} $$ hence by the dominated convergence theorem $$ \lim_{N\to +\infty}\prod_{n=1}^{N}\frac{\left(1+\frac{1}{n}\right)^s}{1+\frac{s}{n}}=\int_{0}^{+\infty}z^s e^{-z}\,dz = \color{red}{\Gamma(s+1)}.$$ You have just re-discovered Euler's product for the $\Gamma$ function.

Answer 2

It is not convergent (unless $s=1$ of course). The key facts are that $$\frac{1+1/n}{1+s/n} = 1 + (1-s)/n + O(1/n^2)$$ and the infinite product $\prod_{n} a(n)$ converges (to a nonzero number) iff $\sum_n (a(n)-1)$ converges.

Answer 3

Hint: $(1+z)^s = 1+sz +O(z^2)$ as $z\to 0.$ (I believe the product converges for $s\notin \{-1,-2,\dots \}.$)