(– 1)"(x+2)" What is the interval of convergence for the series 2 n=1 О.[-6,2) O b.(- 6,2] Oc[-6,– 2] O d. (-6,2)

Find the interval of convergence of the series -1 n8" O (-1, 1) O (-8, 8] O (-8, 8) O (-8, 8) O [-8, 8] O [-1, 1)

O)(4a + 8) i+ (-8a + 24) j, ii) 0 If the interval of convergence of the power series (-1)"(x – c)" is (k, 15 for some k, then find c. 2" In n n=2 Türkçe: Eğer (-1)"(x – c)" kuvvet serisinin 2" In n n=2 yakınsama aralığı bir k sayısı için (k, 15] şeklinde ise, c değerini bulunuz. O 13,00 O 4,75 O 7,50 O -9,00 O -3,50 O 5,67

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.