Question

Find the radius of convergence, R, of the series. 0 (-1)"x" Σ Vn n = 1 R = 1 Find the interval, I, of convergence of the series. (Enter your answer using interval nota I = (-1,1) X

Answer 1

BW ut ult-) Soln q (-1)" xn - n) s/ile n=1 Formula Om Radius of Convergence (R) = lim nya anti Here an (-1) 11 - Hup nti (-1). (MH) n am R anti n (-1). n R well = nh00 (-1)NH (14) 5 (NH) will = 461-) fil Hull n100 m$ (1++)$ onts (-1) = will ntoo lim (i++) lim (1+) (1) ราOC) = olu

at R = (1+4)* (1+0) $ st 11 = So Raelius of Convergence = R = Interval of Convergence Note :- of anxa converges (1) for every 1x)<R (2) The behavious of a power series at 1x1 = R , depomas entirely upon the character of the sedyence {9n3 of its coefficiente

interval of convergence 1x1 < R. 1x1<I [: R=1] Now check for 1x1 = R 1x1 = 1 x = -1,1 (1) fröst check for x= -1 (-1973 nal ni but x=-1 an 2 -7*" 3-1449 = ni ni nal nal => 8 (-12) 2 - - nis nal mal is divergent by p-sesies Test nis n=1 su Note: - p-series np ME) lil sesies is convergent for Þ> 1 lii series is divergent for p <1

So X=-1 , seues is divergent, X=1 * Now check for (-1)^** no but X=1 in series E (-1) lin ni e com nis ne) 8W (-1) nis is convergent by lebnitz Test ne) Note :- Lebnitz-Test :- The alternuting sesies 9 -92 +93-94f --(-199n ton Como Hnen) Converges if each term is numerically less then the preceeling term am> anti tn anel Lim an=0 n00 Hese an= nits lim mo ni it So sesies E n=1 ns is Convergent by lebnitz - Test. So intesual of Convergence is 1x1 and x=1

intesvel of convergence (I) 1x1 <1 and x=1 = (1, or -1 < x < 1 = (-1, H please Reste my answer! Thank you!