Question

number 4

1. Find the limit of the following sequences (find lim an) n n a.) an = n +3 b.) an = V35n n- 2. Determine whether the following series converge or diverge. -3 (n + 2)n + 5 b.) tan-'(n) n2 + 1 a.) 5 nel 3. Determine the radius of convergence and the interval of convergence of the series 2" (x – 3)" n n=1 n=0 (-1)", 2n 4. Using the power series cos(x) (2n)! (-« < x < 0o), to find a power series for the function f(x) = sin(x) sin(3.c) and its interval of convergence. 5. Find the power series representation for the function f(x) = and its interval (3.x - 2)2 of convergence. 6. Find the power series representation for the function f(x) = (1 - 2x)3 and its interval of convergence. 2.c 7. Find a power series for the function f(x) = 3x2 - 52 - 2 and its interval of convergence. 8. Given f(x) = ln(1 – 3x), without using any well known power series, find the MacLaurin series of f(x) and its interval of convergence.

Answer 1

Here we use some trigonometric identity and ratio test to derive the interval of convergence.

Goven Coen = MB (1) 220 (-0. (2n)! (- ocneed h=0 Want to find function a power series for the f(n) = Rin (n) Reni (n) Ren(3) Now ar we know f(n)= ein (n) deri 3n 2 sin n rin 3n Since 2 rin a. RunB = cox (a-) – Colla+B) Thus, f(x)= 1 [cex (1-3o) (n+3n) = { [ con (-2n) - Cox(49) COS

2.1 f(n) = { [ COR 211 – cos an] ( .: COR (-A)= SORA) We are given that (olncool CORM= E CI)? 25 (29)4 n0 Now replacing n .by an, we have Cor an = į G) (22) 24 (2017 h=0 11 2 8 (1) 22 n2n h=0 (2n)! Similarity 8 W Cor 3n = con 32 n2n ho (2n) 1

Now f(n) = Cur an- Cor un 4n] = -ام E GU)"22man n=0 (2A)! Gigh 32 n2n (2n)! ho Σ 2n - ایم (229- 32m) (2n)! n=0 f(x) = 1 ร (42 gh) (-1) 22 2 n=0

Apply the ratio test! fcun+y[4**- genti] x2 (nt) (2n): limeup (26+1) + N2n (-1)" [40 g ki ht Ini eup hto m2. [ant-ghan]. (2n)! [47-Gm] (2n+2) (2n+1821) LI 1+1 nti g n² lim sup KA 2(n+1) (2n+1) g[(49)".] [(119) + n² lim sup LO noo 2(n+1)(2n+1) Tn? <lim eup. 2 (n+1)(2n+1)(419)?] 9 [(419) h* noo

Since 419 L1 lim eup (419 ht' - lim eup (419) = 0. hea te 1/2 lim sup 2 (n+1)(ant) 102 Lo le o Lnco.

This the power series of f(x) converges everywhere.