The expression $\prod_{n=1}^{\infty}(1 - \frac{1}{3n + 2})$ what is the value of the product if $n$ varies from $1$ to $\infty$.
My approach:
I took the log of each term
making it $\sum_{n=1}^{\infty}log(1 - \frac{1}{3n+2})$ so the original product becomes
$\prod_{n=1}^{\infty}(1 - \frac{1}{3n + 2})$ = $e^(\sum_{n=1}^{\infty}log(1 - \frac{1}{3n+2})^)$
now $\frac{1}{3n+2} < 1 \space \forall \space n \space \ge1 $
Now $log(1 + x) \approx x$ so we can write $log(1 - \frac{1}{3n+2}) \space \approx \frac{-1}{3n+2}$ taking limit of which will turn out to $-\infty$ and thus the product should converge to $0$.
Am I correct?
Yes, correct, but could be more precise.
For example, use $1+x\le e^x$, so that $\log(1+x)\le x$ for $x>-1$, thus $$\log(1-\frac1{3n+2})\le -\frac1{3n-2}$$ so their sum indeed converges to $-\infty$.