Show that $\prod _{_{r=1}}^{\infty }\left[1+\frac{x}{r}\right]e^{-x/r}$ converges for all finite x (except for the zeros of 1+x/r) Background: This is Arfken et all Mathematical methods 12.5.8

There is a hint to write the nth factor as $1+a_n$

An infinite product $\prod^\infty(1+a_n)$ diverges or converges with $\sum (a_n)$. Therefore the first term will diverge by comparison with the harmonic, and the second term will converge but I don't know how to proceed with the combined product. 
Also, what is meant by the zeros of 1+x/r?
For $r&gt;|x|$ take logarithms
$$\ln\left[\left(1+\frac xr\right)e^{-x/r}\right]
=-\frac xr+\sum_{n=1}^\infty(-1)^n\frac{x^n}{nr^n}=-\frac{x^2}{2r^2}
+O((x/r)^3).$$
This means that the sum of $\ln\left[\left(1+\frac xr\right)e^{-x/r}\right]$ over all these $r$ is bounded by a multiple of the convergent
series $\sum1/r^2$, so it converges.
Stealing the credit off of Lord Shark's answer and putting it in a form that I understand: Instead of taking the ln just expand the exponential into it's series and multiply the product out into a sum:

$$\prod _{_{n=1}}^{\infty }\left[1+\frac{x}{n}\right]e^{-x/n}=(1+\color{red}{\sum_n \frac{x}{n}}+\sum \frac{x^2}{n_1 n_2}\dots)(1-\color{red}{\sum\frac{x}{n}}+\sum\frac{x^2}{2!n^2}\dots)$$
The red terms in the product cross multiply with 1 and sum to zero, so we get a sum of convergent sums: $$(1+\sum\frac{x^2}{2!n^2}+\sum \frac{x^2}{n_1 n_2}\dots)$$ with each of these sums and higher orders converging with comparison to $\sum\frac{1}{n^2}$ so the final answer that they converge was correct.