_Monotonia
an=1+1/2²+1/3²+...+1/n²
an+1=1+1/2²+1/3³+...+1/n²+1/(n+1)²
an+1-an=1/(n+1)²>0=>
an+1>an=>
an sir monoton crescator
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1≤1
1/2²<1/1*2=1-1/2
1/3²<1/2*3=1/2-1/3
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1/n²<1/n-1/(n+1)=1/(n-1)-1/(n+1)
--------------------------Se aduna
an=∑1/k²=1+1-1/2+1/2-1/3+...+1/(n-1)-1/(n+1)=
2-1/(n+1)=(2n+2-1)/n+1)=(2n+1)/(n+1)→2
Aplici teorema, daca un sir e marginit si e crescator , atunci limita sa este marginea sa superioara.In cazul de fata 2.
Explicație pas cu pas:
