# Question 1

Let $$a_1 = 1$$ and $$a_{n+1} = (n + 1)(a_n + 1)$$. Find $\prod_{n=1}^\infty\left(1+\frac{1}{a_n}\right).$

## Solution by Hobbit & Hercules

### Preamble

#### Definition 1

We define $$T : \mathbb{N} \rightarrow \mathbb{R}$$ by the formula $T(n) = \sum_{k=0}^n \frac{1}{k!}.$

#### Definition 2

As in the question, we define $$a_1 = 1$$ and $$a_{n + 1} = (n + 1)(a_n + 1)$$ for all $$n > 0$$.

#### Definition 3

We define $$P : \mathbb{N} \rightarrow \mathbb{R}$$ by the formula $P(n) = \prod_{k=1}^n \left(1 +\frac{1}{ a_n}\right).$

#### Proposition 1

$P(n) = \frac{a_{n + 1}}{(n+1 )!}.$

#### Proof

Let's consider a single factor of our product series, $$1 + \frac{1}{a_n}$$. Note that $$1 + \frac{1}{a_n} = \frac{a_{n} + 1}{a_n}$$. From Definition 2 we know that $$a_{n + 1} = (n + 1)(a_n + 1)$$ which implies that $$a_n + 1 = \frac{a_{n+1}}{n+1}$$. We can then substitute this into $$\frac{a_{n} + 1}{a_n}$$ in order to obtain $$1 + \frac{1}{a_n} = \frac{a_{n+1}}{(n+1)a_n}$$.

Now let's consider our product series using what we have learnt about each factor: $P(n) = \left(1 + \frac{1}{a_1}\right)\left(1 + \frac{1}{a_2})(1 + \frac{1}{a_3})\cdots(1 + \frac{1}{a_n}\right).$ Applying our substitution for each factor we can write this in the form $P(n) = \frac{a_2}{2 a_1} \frac{a_3}{3a_2} \cdots \frac{a_{n+1}}{(n+1)(a_n)}$ and by shifting each denominator to the left we get $P(n) = \frac{1}{2a_1} \frac{a_2}{3a_2} \frac{a_3}{4a_3}\cdots\frac{a_n}{(n+1)(a_n)} a_{n+1}.$ At this point we can cancel out factors that appear in both the numerator and denominator to obtain $P(n) = \frac{1}{1}\frac{1}{2}\frac{1}{3}\cdots\frac{1}{n}\frac{1}{(n+1)} a_{n+1} = \frac{a_{n+1}}{(n+1)!}.$

We can now announce our solution:

### Main solution

$\prod_{k=1}^\infty \left(1 + \frac{1}{a_n}\right) = e.$

#### Proof

Consider the functions $$P(n)$$ and $$T(n)$$. We will prove by induction on $$n$$ that $$P(n) = T(n)$$ for all $$n$$. We will then deduce that both functions have the same limit as $$n \rightarrow \infty$$, namely $$e$$.

#### Proposition 2

$P(n) = T(n).$

#### Proof

This is a proof by induction on $$n$$:
Base case:
We will show that $$P(1) = T(1)$$. We have $$P(1) = (1 + \frac{1}{1}) = 2$$ and $$T(1) = \frac{1}{0!} + \frac{1}{1!} = 1 + 1 = 2.$$ Hence, $$P(1) =T(1) = 2.$$
Inductive step:
We will assume that $$P(n) = T(n)$$. We want to show that $$P(n+1) = T(n+1)$$. We know that $P(n+1) = P(n)(1 + \frac{1}{a_{n+1}})$ and $T(n+1)= T(n) + \frac{1}{(n+1)!}.$ But we know from Proposition 1 that $$P(n) = \frac{a_{n + 1}}{(n+1 )!}$$. So $P(n + 1) = \frac{a_{n + 1}}{(n+1 )!}(1 + \frac{1}{a_{n+1}}) = \frac{a_{n + 1}}{(n+1 )!} + \frac{1}{(n+1)!} = P(n) + \frac{1}{(n + 1)!}.$ But we assume that $$P(n) = T(n)$$ which implies that $P(n+1) = T(n) + \frac{1}{(n+1)!} = T(n+1),$ as required.

Therefore $$P(n) = T(n)$$ for all positive natural numbers. This concludes Proposition 2.

Consequently, because $$P(n) = T(n)$$ for all positive natural numbers $$n$$ we conclude that they must have the same limit as $$n \rightarrow \infty.$$ Therefore, since $\lim_{n \to \infty} T(n) = e,$ we know that $\lim_{n \to \infty} P(n) = \prod_{k=1}^\infty (1 + \frac{1}{a_n}) = e.$

#### Remarks on Question 1

• This problem raises the interesting question "Can we write other Maclaurin and Taylor series as infinite products in this way?" For example, is it possible to write a number such as $$ln(2)$$ in this way?
• Knowing that $\lim_{n \to \infty} P(n) = \prod_{k=1}^\infty (1 + \frac{1}{a_n}) = e,$ we can now calculate $$e^x$$. Indeed, we can think of it as $$\lim_{n \to \infty} \prod_{k=1}^n (1 + \frac{1}{a_n})^x = (e)^x = e^x.$$
• Let $$P : (\mathbb{N},\mathbb{R}) \rightarrow \mathbb{R}$$ be defined by taking $$P(n,x) =\prod_{k=1}^n (1 + \frac{1}{a_n})^x$$. Let $$T : (\mathbb{N},\mathbb{R}) \rightarrow \mathbb{R}$$ be defined by taking $$T(n,x) = \sum_{k=0}^n \frac{x^k}{ k!}$$. Let $$G : (\mathbb{N},\mathbb{R}) \rightarrow \mathbb{R}$$ be defined by taking $$G(n,x) = (1 + \frac{x}{n})^n$$. By way of example, the graph below indicates what value of $$n$$ is needed in order for $$P(n,2)$$ (in green), $$T(n,2)$$ (in blue), and $$G(n,2)$$ (in red) to get close to their limiting value $$e^2$$: We conjecture that for all $$x > 1$$, we have $$P(n,x) > T(n,x) > G(n,x)$$.