# Find the limit?

## a_1=1 & a_n=n(a_(n-1)+1) for n=2,3,..... Define ${P}_{n} = \left(1 + \frac{1}{a} _ 1\right) \cdot \left(1 + \frac{1}{a} _ 2\right) \cdot \ldots \ldots \ldots \ldots \ldots . \cdot \left(1 + \frac{1}{a} _ n\right)$ Then ${\lim}_{n \rightarrow \infty} {P}_{n}$ is?

Apr 10, 2018

${\lim}_{n \to \infty} {P}_{n} = e$

#### Explanation:

Note that:

${a}_{n + 1} = \left(n + 1\right) \left({a}_{n} + 1\right)$

${a}_{n + 1} / \left(n + 1\right) = {a}_{n} + 1$

so:

$\left(1\right) \text{ } 1 + \frac{1}{a} _ n = \frac{{a}_{n} + 1}{a} _ n = \frac{1}{n + 1} {a}_{n + 1} / {a}_{n}$

Then, having defined:

${P}_{n} = {\prod}_{k = 1}^{n} \left(1 + \frac{1}{a} _ k\right)$

we have:

${P}_{n} = {\prod}_{k = 1}^{n} \frac{1}{k + 1} {a}_{k + 1} / {a}_{k}$

P_n = 1/((n+1)!)prod_(k=1)^n a_(k+1)/a_k

P_n = 1/((n+1)!)( (a_(n+1)/a_n)(a_n/a_(n-1))...(a_3/a_2)(a_2/a_1))

P_n = a_(n+1)/((n+1)!)

Now if we substitute the formula for ${a}_{n + 1}$ we get:

P_n = ((n+1)(a_n+1))/((n+1)!)

P_n = ((a_n+1))/(n!)

P_n = a_n/(n!)+1/(n!)

P_n = P_(n-1)+1/(n!)

Let now:

P_0 =1 = 1/(0!)

so that:

P_1 = 1+1/a_1 = P_0 +1/(1!)

then:

P_2 = P_1 +1/(2!) = sum_(k=0)^2 1/(k!)

and clearly we can establish by induction that:

P_n = sum_(k=0)^n 1/(k!)

and then:

lim_(n->oo) P_n = sum_(k=0)^oo 1/(k!)

Remembering that the MacLaurin formula for the exponential function is:

e^x = sum_(k=0)^oo x^k/(k!)

converging for every $x \in \mathbb{R}$ we can see that for $x = 1$:

sum_(k=0)^oo 1/(k!) = e

and conclude that:

${\lim}_{n \to \infty} {P}_{n} = e$