La Moneda

1) n=1∞n+13n+1
Usando el criterio de la razón: si r <1 (convergentesi r >1 divergentesi r=1 criterio no aplica

r= an+1an
Entonces
an+1= n+1+13n+1+1 = n+23n+3+1 = n+23n+4
an= n+13n+1Luego
r = Lim n+23n+4n+13n+1 = r= Lim n+2 . 3n+1n+1 . 3n+4
n ∞ n ∞

r = Lim 3n2+n+6n+33n2+4n+3n+4 = Lim 3n2+7n+33n2+7n+4n ∞ n ∞

r = Lim n23+7nn2++3n2n23+7n2+ 4n2 = Lim 3+7n+3n22+7n+4n2
n ∞n ∞

r= 3+7∞+3∞23+7∞+4∞2 = 33 = 1

2) n=1∞ne-n2

Usando el Criterio de la Integral:

I= Lim oLaxdx ; an= ar= xe-x2
L ∞

I= Lim oLaxdx; an=ar=xe-x2
L ∞

I= Lim oLxe-e-x2dx U= -x2du=-2xdx → -du2 = xdx
L ∞

I= Lim -12 0Le2du = I= Lim -12 e-x2 LoL ∞ L ∞

I= -12 . Lim e-L2 - eo
L ∞

I= -12 . Lim e-L2- 1
L ∞

I= -12 e-∞ + 12 El Limite Existe

Entonces n=1∞ne-n2 → Converge

3) n=1∞14n

Usando el Criterio de la razón;

an= 14n; an+1= 14n+1

r= Lim an+1an = Lim 14n+114n = Lim 4n4n+1
n⇢a n⇢a n⇢a

r= Lim 4n4n . 4 = 14 . Lim 4n4 = 14 L1n⇢a n⇢a

N=1∞14N ⟶Coverge
4) n=1∞(n+12n+1)n

Usando el Criterio de la Raíz:

An= n+12n+1n

Lim nan = Lim nn+12n+1n = Limn+12n+1
n⇢ ∞ n⇢ ∞ n⇢ ∞

Lim n1+1nn2+1n = Lim 1+1n2+1n
n⇢ ∞ n⇢ ∞

Lim 1+1∞2+1∞ = 12 L1
n⇢ ∞...