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

Answer:

#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#