I need help with a textbook exercise (Gamelin's Complex Analysis, Ch.8 Section 3 Problem 14). This exercise requires me to show that $$\lim_{n\to\infty} \prod_{-m\leq k \leq tm} \left(1+\frac{z}{k}\right) =\frac{\sin \pi z}{\pi z} t^z.$$
Since we know that $$\frac{\sin \pi z}{\pi z}= \prod_{k=1}^{\infty}\left(1-\frac{z^2}{k^2}\right)$$ I have tried factoring this expression and massaging it into the above form. However, I don't know how to include the t factor into the equation. Also the term isn't symmetric so the negative product doesn't easily come forth. How would one go about solving this? Is the Gamma function required?
For a solution, it's sufficient to use $\lim\limits_{n\to\infty}\sum\limits_{an<k<bn}\frac1k=\log\frac ba$ for $0<a<b$, which may be shown in different ways (Riemann sums, or the fact that $\sum\limits_{0<k<n}\frac1k-\log n$ has a finite limit as $n\to\infty$, etc.).
Assume first that $t>1$, then we need to prove that $\lim\limits_{n\to\infty}\prod\limits_{n<k<tn}(1+z/k)=t^z$.
But, indeed, since $\log(1+z/k)=z/k+O(1/k^2)$ as $k\to\infty$, we get $$\sum_{n<k<tn}\log\left(1+\frac zk\right)=z\sum_{n<k<tn}\frac1k+O(n)\cdot O(1/n^2)$$ as $n\to\infty$, and the limit equals $z\log t$, as expected.
For $0<t<1$ we prove that $\lim\limits_{n\to\infty}\prod\limits_{tn<k<n}(1+z/k)=t^{-z}$, the same way.