$\lim_{a \to \infty}\prod_{i=0}^{a} x_i = \infty$ for $x_i > 1$?

Question

For $x_i > 1$, where $i$ is an index and $x_i$ real number, $\lim_{a \to \infty}\prod_{i=0}^{a} x_i = \infty$ always? How does one prove/disprove this?

Answer 1

It is relatively easy to find a counterexample showing that this is not true. Simply choose any increasing bounded sequence $(p_n)$ such that $p_0=1$. I.e., you have $p_n<p_{n+1}$ for each $n$ and, since the sequence is bounded, there exists a finite limit $\lim\limits_{n\to\infty} p_n=P$.

Now you can put $x_0=1$ and $$x_n = \frac{p_n}{p_{n-1}}$$ for $n>1$.

In this way you have $$\prod_{k=0}^n x_k = p_n$$ and $$\prod_{k=0}^\infty x_k = \lim\limits_{n\to\infty} \prod_{k=0}^n x_k = P.$$

Moreover, you have $x_k>1$ for each $k$.

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It was already mentioned in another answer that there is a general criterion for convergence of infinite products of the form $\prod\limits_{k=0}^\infty (1+a_k)$ (where $a_k$ is a real sequence). You will probably find several posts about this if you browse (or search) the infinite-product tag a bit. For example:

• sufficiency and necessity of convergence of $\sum a_n$ wrt convergence of $\prod (1 + a_n)$
• $\prod (1+a_n)$ converges iff $ \sum a_n$ converges
• Suppose $a_n>0$ for $n\in \mathbb{N}$. Prove that $\prod_{n=1}^\infty (1+a_n)$ converges if and only if $\sum_{n=1}^\infty a_n<\infty$.

Answer 2

If you write $x_i=1+u_i>1$ then the product converges iff $\sum_i u_i <+\infty$. You may see this by taking log and make a comparison through Taylor