Why does $\prod_{n=1}^\infty \frac{n^3 + n^2 + n}{n^3 + 1}$ diverge?

Question

$\prod_{n=1}^\infty \frac{n^3 + n^2 + n}{n^3 + 1}$ diverges, and I have no idea why? It would seem using L'hop, $\frac{n^3 + n^2 + n}{n^3 + 1}$ goes to 1. So it should end up just being $1\cdot1\cdot1\cdots$, which makes me feel it converges. Is the problem that it always adds that $.000\ldots0001$?

Answer 1

When you have a product $\prod_n A_n$, the condition that $A_n \rightarrow 1$ is necessary but not sufficient. It is in fact equivalent to the fact that when you have a sum $\sum_n B_n$, the condition that $B_n \rightarrow 0$ is necessary but not sufficient. Thus the intuition is the same: just as the terms of an infinite sum may approach zero "too slowly", the terms of an infinite product may approach $1$ "too slowly". The correspondence can be derived by taking logarithms:

$$ \log \prod_{n=1}^k A_n = \sum_{n=1}^k \log(A_n)$$

In particular, for your problem

$$ \log \prod_{n=1}^k \frac{n^3 + n^2 + n}{n^3 + 1} = \sum_{n=1}^k \log\frac{n^3 + n^2 + n}{n^3 + 1} $$ and you can check that the above series diverges as $k \rightarrow \infty$.

Answer 2

Divergence of a product of nonzero terms $$\prod_{k=1}^{\infty} a_k$$ is equivalent to divergence of the sum $$ \sum_{k=1}^{\infty} \log{a_k}$$ In your case, the corresponding sum diverges.

Answer 3

Intuitively, as $\frac{n^3 + n^2 + n}{n^3 + 1} \sim 1$, if we erase $n$ from the top and $1$ from the bottom, the fraction should get smaller. Lets prove it:

$$\frac{n^3 + n^2 + n}{n^3 + 1}-\frac{n+1}{n}=\frac{n^2-n-1}{n(n^3+1)}$$

Thus, for $n \geq 2$ we have

$$\frac{n^3 + n^2 + n}{n^3 + 1} > \frac{n+1}{n}$$

Now, since it is telescopic:

$$\prod_{n=1}^N \frac{n+1}{n} =N+1$$

Answer 4

Notice that for $n\ge1$ you have $$\frac{n^3+n^2+n}{n^3+1}= 1+\frac{n^2+n-1}{n^3+1} \ge 1+ \frac{n^2-n+1}{n^3+1} = 1+\frac1{n+1} = \frac{n+2}{n+1}.$$

Therefore $$\prod_{k=1}^n \frac{k^3+k^2+k}{k^3+1} \ge \prod_{k=1}^n \frac{k+2}{k+1} = \frac32 \cdot \frac 43 \cdots \frac{n+2}{n+1} = \frac{n+2}2.$$

And for $n\to\infty$ we have $\lim\limits_{n\to\infty} \frac{n+2}2=\infty$.