Is $\prod_{i=1}^{\infty}(1+ \frac 1 {8i})=\infty$?

Question

As a lemma for a certain analysis problem, I want to show that the infinite product $\prod_{i=1}^{\infty}(1+ \frac 1 {8i})$ diverges to $\infty$.

Having computed some $\prod_{i=1}^{n}(1+ \frac 1 {8i})$ terms in Wolfram, I am pretty sure that the series diverges but I'm not sure how to show it rigorously.

Clearly $\prod_{i=1}^{n}(1+ \frac 1 {8i})\geq (1+ \frac 1 {8n})^n$ but this doesn't really help because I'm pretty sure that the RHS converges. (I know that $(1+ \frac 1 n)^n$ converges to $e$.)

Answer 1

It diverges.

When you want to verify if an infinite product of (positive) numbers converges or diverges, what it takes is just to look what happens with the logarithms:

$\log\left(\prod_{i=1}^\infty \left(1+\frac{1}{8i}\right)\right) = \sum_{i=1}^\infty \log\left(1+\frac{1}{8i}\right)$

Using the standard inequality $\log(1+x)\geq \frac{x}{x+1}$ that holds for $x\geq -1$, one gets:

$\log\left(1+\frac{1}{8i}\right) \geq \frac{\frac{1}{8i}}{1+\frac{1}{8i}} = \frac{1}{8i+1}$.

In particular:

$$\log\left(\prod_{i=1}^\infty \left(1+\frac{1}{8i}\right)\right) = \sum_{i=1}^\infty \log\left(1+\frac{1}{8i}\right)\geq \sum_{i=1}^{\infty} \frac{1}{8i+1}$$

And the RHS is clearly divergent, since it is essentially a harmonic series (use the $p$-criterion for $p=1$).

Answer 2

Perhaps the easiest way to prove this is to note that $$\prod_{i=1}^n\left(1+\frac1{8i}\right)\ge \sum_{i=1}^n\frac1{8i}$$ since after opening all brackets we see that the lefthandside is equal to the righthandside plus some other positive terms. Now pass to the limit with $n\to \infty$.