Question

Criteriul clestelui. Sa se calculeze limitele sirurilor definite prin termenul general. ​

Answer 1

Răspuns:

1/√(n²+1)<1/√n²<1/n

1/√(n²+2)<1/√n²<1/n

....................................

1/√(n²+n)<1/√n²<1/n

Le adui termencu termen

1/√(n²+1)+1/√(n²+2)+....+1/√(n²+n)<1/n+1/n+...1/n=n/n=1 =an  an->1

xn≤an

Dar

1/√(n²+1)>1/√(n²+n)

1/√(n²+2)>1/√(n²+n)

............................................

1/√(n²+n)≥1/√(n²+n)

Le adui

1/√(n²+1)+1/√(n+2)+...+1/√(n²+n)≥1/√(n²+n)+1/√(n²+n)+...1/√(n²+n)=n/√(n²+n)=

n/n√(1+1/n)->1 =bn  bn->1=>

bn≤xn≤an

1≤xn≤1=>

lim xn=1

Explicație pas cu pas: