Find the limit?

a_1=1 & a_n=n(a_(n-1)+1) for n=2,3,..... Define P_n=(1+1/a_1)*(1+1/a_2)*................*(1+1/a_n)

Then lim_(nrarroo)P_n is?

1 Answer
Apr 10, 2018

lim_(n->oo) P_n = e

Explanation:

Note that:

a_(n+1) = (n+1)(a_n+1)

a_(n+1)/(n+1) = a_n+1

so:

(1) " " 1+1/a_n = (a_n+1)/a_n= 1/(n+1) a_(n+1)/a_n

Then, having defined:

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

we have:

P_n = prod_(k=1)^n 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 RR we can see that for x=1:

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

and conclude that:

lim_(n->oo) P_n = e