Question

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[03] Letan = n3"x", then lim lan+1/an] = (A)x/31; (B) 3xl; (C)|3/x]; (D) xl; (E) None of these. [04] The radius R of convergence of n3yn is (A) 1; (B) 72; (C) 3; (D) 1/3 (E) None of these. [05] The interval of convergence of Zn3"x" is (A) [-1/4,44); (B) (-1/4, 14); (C) (-1/4,44]; (D) [-1/4,44]; (E) None of these. [06] The radius of convergence ofx/n3") is (A) 1; (B) 2; (C) 3; (D) 4; (E) None of these. [01] The radius of convergence of (x-1)" Vn is (A) 2; (B) 1; (C) 2; (D) 3; (E) None of these. _ [04] Eg"/n! is (A) convergent; (B) divergent; (C) None of these. [05] 2 [(2n)/(n2+2n+2) is (A) convergent; (B) divergent; (C) None of these. [02] 31/n! is (A) conditionally convergent; (B) absolutely convergent; (C) divergent; (D) None of these. (06) 1/(n2+1)< (A) 1/(n2-1); (B) 1/n2; (C) 1/(n2-2); (D) 1/(2n²); (E) None of these.

Answer 1

3) Ean=5nnxn we have an=namach Ponti (19+1) 39+'cnti =(1+1)3" 31032 Then lim I antil = lim an nyo 1 (n+1337.3 (869.a nga But 1X11+1) 3.0 - lim nto I = 1 (1+0): 321 = 138 Be civen power Series power Series Enam och {anco-as) we have wher ar centre- - +) ana nahi an #1 = (n+1) 31+1 kwow that JU we Radius of con RI convergen (R) & Ratio Test b= lim antill alim ((n+1) 8m 3 ho I an n tool nga a ten Mic ) 11 2 lim n too

- E o = (1+oja = 3 ROBO REŠ Hence Rading of convergent (R)= 1/2 ou The Interval of convergence of Enghoch :- o we have cx-ojn ie centre = 0 and Rad we know that Interval of convergence. -R< x-aLR ) - <«-02 Ś > - < x < į. Now check at endpoint out x = -1/3 in ① we get EM 3".(-3455 - 59.384. Cryo Ec-1".n to - dot again put a = 1/3 in o we get Eno 3" ( 13 14 = Enis it. En to Honce interval of comergence. *<< 5) (* ) . © Alone of these 257 Lommeleg

6 Criven power series Exe" (n3") som { to t we have j anti = intus g9+1 I have an= ulan an= 2.3n. Then by Ratio Test we know that 1 = lim lant nto | 3 um II n tool in til 30. a * "20, tn 3x1 (1+0)3 Que = litos3 Š. 7 ☆ = ☆ R =3. Hence Radius of convergence CRJ =3. @ brinen power series. E6-1) Ja we have an= coce-dd Jn. anti : snui Then by "Ratio Test

= lot more J 14 m 1 - Nitoas Itin s1. . Hence Rading of convergoce (R321. Wie +20+2 > 5[1212)} *2012 ]": E[ mm] X2, we have Itzi + VA Then by Root test ! - linn n- nyo bas 2 toto i lim lan| kn = 2 1 nta Thon by Root Test given series is divergent - One

0 539/! Then by Ratio Test lims | ant la limo lenta lantil ein 8. 3 x n in l ndoo | lim hto huco nto 1 an lim. 1 =3 > Then by Ratio Test given Out series is divergent. b2+1