Have you seen an infinite product before?

First, let's dispel some misconceptions. One may first notice that this looks like a telescoping product because

$\prod_{n=1}^{\infty} \frac{n}{n+1} = \frac{1}{2} \frac{2}{3} \frac{3}{4} ...$

And because the numerator of one fraction cancels with the denominator of the previous fraction, the product must converge to 1 since it's the only numerator that cannot cancel with any denominator. However, this is incorrect. Let's revisit what it means for an infinite product to converge by first looking at a short discussion of convergence of infinite series.

An infinite series $\sum_{n=1}^{\infty} a_n$ will converge if the sequence $\{s_k\}_{k=1}^{\infty}$ converges ( where $s_k = \sum_{n=1}^{k} a_n$ ). We call each $s_k$ a partial sum and we call $\{s_k\}_{k=1}^{\infty}$ the sequence of partial sums. Notice that a partial sum simply sums the first $k$ terms of a series. We can succinctly say that an infinite series converges if the sequence of its partial sums converges.

Furthermore, if this sequence of partial sums converges, then we define the sum of the series to be the value that this sequence converges to. For example, let's look at

$\sum_{n=1}^{\infty} \frac{1}{2^n}$

So let's start by looking at the partial sums of this series

$s_k = \sum_{n=1}^{k} \frac{1}{2^n} = \frac{2^k - 1}{2^k} = 1 - \frac{1}{2^k}$

Note that I found a closed form expression for this sum by using the formula for the sum of finitely many terms of a geometric series. So this means our sequence of partial sums is

$\{1 - \frac{1}{2^k}\}_{k=1}^{\infty}$

And from this, we see that since

$\lim_{k \to \infty} 1 - \frac{1}{2^k} = 1$

We conclude that the sequence of partial sums converges to $1$ and therefore

$\sum_{n=1}^{\infty} \frac{1}{2^n} = 1$

Well you now may be asking, "but what does this have to do with infinite products?" Well it terms out the definition of the sum of an infinite series is very similar to the definition of the sum of an infinite product. Instead of looking at partial sums, we look at partial products.

An infinite product $\prod_{n=1}^{\infty} a_n$ will converge if $\{p_k\}_{k=1}^{\infty}$ converges (where $p_k = \prod_{n=1}^{k} a_n$ ). We call $p_k$ a partial product and we call $\{p_k\}_{k=1}^{\infty}$ the sequence of partial products. Similar to infinite series, partial products simply multiply the first $k$ terms of the infinite product and the infinite product converges when the sequence of partial products converge. The product of the infinite product is defined to be the value of which the sequence of partial products converges to. Looking at the product in my question,

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

We can find the partial product

$p_k = \prod_{n=1}^{k} \frac{n}{n+1} = \frac{1}{2} \frac{2}{3} ... \frac{k}{k+1} = \frac{1}{k+1}$

The terms telescope in this partial product to reduce to $\frac{1}{k+1}$ . So this means the sequence of partial products is

$\{ \frac{1}{k+1}\}_{k=1}^{\infty}$

Now, we take the limit:

$\lim_{k \to \infty} \frac{1}{k+1} = 0$

We see that the sequence of partial products converges to $0$ and thus,

$\prod_{n=1}^{\infty} \frac{n}{n+1} = 0$

As desired.

---

This is my first time writing a solution in a long time. I looked for typos and didn't see anything. Please comment if I have made any mistake or if anything is unclear! Thanks

-Varun Gudibanda

Applying telescopic series $\displaystyle \prod_{n=1}^{\infty} \frac{n}{n+1} =\lim_{n \to \infty} \frac{1}{n+1}=0$