Răspuns:
∑1/(n²-1)=∑1/(n-1)(n+1)=
1/1*3+1/2*4+1/3*5+....1/(n-1)(n+1)=
1/2(1/1-1/3)+1/2(1/2-1/4)+1/2(1/3-1/5)+...+1/2(1/(n-1)-1/(n+1))=
1/2(1-1/3+1/2-1/4+1/3-1./5+....+1/(n-1)-1/(n+1))=1/2[3/2-1/(n+1)]=
3/4-1/4(n+1)<1
sirul sumelorpartiale este conveergent, rezulta seriaeste convergenta
1/4(n+1)→0=> seria tinde la3/4
Explicație pas cu pas:
