Question

prove limit n - > infinity { {n+1) (n+2) (n+n)^1/n} /n = 4/e​

Answer 1

$\large\bf{\underline\orange{Answer \:↝}}$

(1/n)*[(n+1)(n+2).....(n+n)]^(1/n)

= [(1+1/n)*(1+2/n).....(1+n/n)]^(1/n)

lim n-inf [(1+1/n)*(1+2/n).....(1+n/n)]^(1/n)

= lim n-inf exp{ (1/n) log ((1+1/n)*(1+2/n).....(1+n/n))}

Now log(1+k/n)= k/n - (1/2)*(k/n)^2+(1/3)*(k/n)^3.............

and log (a1.a2......an) = log(a1)+log(a2)+......log(an)

log ((1+1/n)*(1+2/n).....(1+n/n)

= k=1:n Σ (j=1:n(Σ ( j)^k/ k)*(-1)^(k)

lim n- inf exp{ (1/n) log ((1+1/n)*(1+2/n).....(1+n/n))}

=exp( Σ coefficient of highest powers of n in the above power series)

= exp( 1/2-1/6+1/12-1/20-1/30+ 1/42+-------)

Now we need to calculate S= 1/2-1/6+1/12-1/20+1/30 - 1/42+---

S= 1/2-1/6+1/12-1/20+1/30 -1/42 +---

= (1/2+1/12+1/42+ ...............) - (1/6+1/20+ 1/42................)

= S1 - S2

S1= 1/2+1/12+1/42...............

= 1-1/2+ 1/3-1/4+ 1/5-1/6 +.............

= ln2

S2 = 1/6+1/20+ 1/42..............

= 1/2-1/3+1/4-1/5+1/6-1/7 .........

= 1- (1/2-1/3+1/4-1/5+.........)

= 1- ln2

S= S1-S2 = ln2-(1-ln2)= 2ln2-1= ln4 -1

lim n-inf [(1+1/n)*(1+2/n).....(1+n/n)]^(1/n)

= exp( 1/2-1/6+1/12-1/20-1/30+ 1/42+-------)

= exp(ln4 -1 )

= exp(ln4)*exp(-1)

= 4/e Hence proved.