The integral tests says that if an=f(n), then the series 2 an is convergent if and only n =1 if the integral J F(x)dx is convergent as long as the function f is BLANK-1, BLANK- 2, and BLANK-3 on the interval X21. BLANK-1 Add your answer BLANK-2 Add your answer BLANK-3 Add your answer .T dx= lim x-2dx= lim -Tl+1¬1= lim +1 = 1 Since the integral converges and therefore the series 2 K=1 K? also converges, and <1+1=2. K=1 K2

Lemma 4. Let {yn} be a positive solution of equation (1.1) with the corresponding sequence {z„} e So for n > N. Then: (i) (1– p)zn N; (ii) {zn¢n} is increasing for all n > N. Proof. Assume that {yn} is a positive solution of equation (1.1) with the corres- ponding sequence {zn} E So. Then z, is positive, Zn 2 yn, and Yn = Zn – Pnyo(n) 2 (1– p)zn, n>N> no, so (i) is proved. It easy to see that z, E So implies lim b,(Az,)“ = 0; n00 otherwise we would eventually have Az, > 0 contradicting z, E So. Similarly, lim a„A(b,(Azn)“) = 0. A summation of equation (1.1) then yields 9szs+1 Zn+1 LIs. s=n s=n s=n Summing once more, we obtain as t=s s=n or Azn 2-Zn+1Qn. Hence, A(zn&n) = PnAzn + Zn+1Ao, > Zn+1(A0n – OnQn) =0 since {0,} is a solution of the difference equation (Ao, – Qnºn) = 0. Therefore, {z,9n} is increasing and this completes the proof of the lemma.