Calculo Multi
Enviado por gwitto • 21 de Octubre de 2012 • 287 Palabras (2 Páginas) • 371 Visitas
SUMAS PARCIALES
Javier navas
Carlos jaimes
Luis Alberto rivera
Hovan pinto herrera
∑▒〖 5/(n(n+1))〗
S1 = 5/2
S2 =5/2+5/6= 10/3
S3 =5/2+5/6+5/12=15/4
Sn =5n/(n+1)
lim┬(n→∞)〖Sn↔ lim┬(n→∞)〖5n/(n+1)〗 〗= (5n/n)/((n^2+n)/n)= (5n/n)/(n^2/n+1/n)= 5/(n/n)= 5/1= 5
La sucesión converge a 5
∑▒〖1/3-1/(n+3)〗
S1= 1/12
S2= 2/15
S3= 1/6
S4= 4/21
Sn= n/(3n+9)
lim┬(n→∞)〖n/(3n+9)= (n/n)/(3n/n+9/n) = 1/(3+9/n)〗 = 1/(3+0) = 1/3
La sucesión converge a 1/3
GEOMETRICAS
∑▒1/3^n
1/(3*3^(n-1) )= 1/3*1/3^(n-1) → 1/3*(1/3)^(n-1)
1/3<1→CONVERGE
∑▒〖3^2n*2^(1-n) 〗
9^n*2*2^(-1)= 2* 9^n/2^n = 2*(9/2)^n = 2* 9/2*(9/2)^(n-1)
9/2>1 →DIVERGE
INTEGRALES
∑▒〖〖3n〗^2 n^3 〗
∫▒〖〖3n〗^2 n^(3 ) dn =〗
u= n^3
du= 〖3n〗^2 dn
∫▒〖u du= u^2/2〗 → (n^3 )^2/2 = n^6/2
∑▒(2n-7 )/(n^2- 7n+10)
∫▒〖(2n-7)/(n^2- 7n-10) dn=〗
u= n^2- 7n-10
du= 2n-7 dn
∫▒u/du=lnu → ln〖n^2- 7n+10
SUMAS PARCIALES
Javier navas
Carlos jaimes
Luis Alberto rivera
Hovan pinto herrera
∑▒〖 5/(n(n+1))〗
S1 = 5/2
S2 =5/2+5/6= 10/3
S3 =5/2+5/6+5/12=15/4
Sn =5n/(n+1)
lim┬(n→∞)〖Sn↔ lim┬(n→∞)〖5n/(n+1)〗 〗= (5n/n)/((n^2+n)/n)= (5n/n)/(n^2/n+1/n)= 5/(n/n)= 5/1= 5
La sucesión converge a 5
∑▒〖1/3-1/(n+3)〗
S1= 1/12
S2= 2/15
S3= 1/6
S4= 4/21
Sn= n/(3n+9)
lim┬(n→∞)〖n/(3n+9)= (n/n)/(3n/n+9/n) = 1/(3+9/n)〗 = 1/(3+0) = 1/3
La sucesión converge a 1/3
GEOMETRICAS
∑▒1/3^n
1/(3*3^(n-1) )= 1/3*1/3^(n-1) → 1/3*(1/3)^(n-1)
1/3<1→CONVERGE
∑▒〖3^2n*2^(1-n) 〗
9^n*2*2^(-1)= 2* 9^n/2^n = 2*(9/2)^n = 2* 9/2*(9/2)^(n-1)
9/2>1 →DIVERGE
INTEGRALES
∑▒〖〖3n〗^2 n^3 〗
∫▒〖〖3n〗^2 n^(3 ) dn =〗
u= n^3
du= 〖3n〗^2 dn
∫▒〖u du= u^2/2〗 → (n^3 )^2/2 = n^6/2
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