Please answer the following question attached below; Information; ∞ f(x) = Σman. n=0 Please find the following from information below; 1) Determine f'(x), reindexing to ensure that the sum begins with n=0 2) For all x on some interval, let f(x) = f'(x) and determine relationship between coefficients of both series which makes this true. 3) From 2)'s relationship and a recursive argument, write general coefficient cn in terms of co (0th order coefficient). Recursive argument begins with c₁ in tern of co then c₂ in terms of co and continuing. Hence, provide f(x) and what function does the series define, given f(x) = f'(x)?

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.

Please make sure the answers are clear n easy to read. Please do it on white paper. Again make sure the answer is clear.  do it on white paper

(2n+ 1)x Please explan Determine whether the series sin- converges or diverg es. If it converges, find its sum. thish L and how it nelotes to the n-D0 (4r + 1)x Since sin =1 and sin =-1 and the angle may be incremente by any multiple of 2x, then sin 2 =1 and (4r+3)a sin =-1, (any r). How does this nelate to the answer. Examine the partial sums of the series. The expanded series is sin 3n 5x + sin + sin ,+ sin + sin+ .. ansver Notice that the angles in consecutive even terms (n is even) differ by 2x, and the angles in consecutive odd terms (n is odd) differ by 2x. Since sin 0 = sin (0+2mx), all the even terms are equal to 1, and all the odd terms are equal to -1. Evaluating each term, the series is 1+(-1)+1+(-1)+1+ The partial sums altermate between 1 and 0, tending to (2n + 1)a neither value. Therefore, lim sin- k00 n=0 does not exist and the series diverges. detail. Pelease explain n thanke gou.