Question

Determine the radius and interval of convergence for the following power series x²k+1 23k 7 MS
Determine the radius and interval of convergence for the flowing series

Answer 1

1) Given series is į(-)* map zet u =(-* Now, en - - * * * * * * * * = H - How the series Dicas e T * convergents in 1. That's te bof <1=\x{<1. Then the Radius of Convergence – sep :3(-1* -sav { \ - then the series ( rukt Then the Radius of Convergence = supx: (-1 7 = sup {1}=1 then the series K + 1) k12 is convergent on (-1,1) Now check at the end points, for r=-1 we have the series (_12k+k+1 = > (+1) k=1 k+1) (k+1) (-1 5. Since lim (k+1) 5 =0 then by Leibnitz test (k+1) k+ we can conclude that the series 2 is convergent at x=-1. +1) (k

Now, for x=1 we also have Now, for x=1 we also have § (-)* (m. = = 0, which is same pretty as above, then similarly by which is same pretty as above, then similarly by Leibnitz test we can conclude that the series ( 4 r2+1 (k+1? Is conve is convergent at x=1. 224+1 Then the interval of convergence of the series > Then the internal of convergence of the series (- * * O * (-1). 3) Given series is Now, in: - - Now the series sonsorgen 4. The Il <1=of <3=}<15. Then the Radius of Convergence = sup (le = sup{13}= 73. Then the series is convergent on r2+1 Then the Radius of Convergence = sup{x}: 3. Then the series 26-1 is convergent on

(-13,13). * x2k+1 Now check at the end points, for x=-3 the series ak-1 becomes kaj 3-1 beca 2:13. godine. 34-354793* - get which ia divergen series. Now, for x = 15 ten eise become si o si wisk i ako diversen weis. Now, for x = V3 then series becomes = 32 which is also a divergent series. 34-1 A8 Then we see that at the end point the series) fails to convergent. Then the interval of the series : Series .r+k+1 kel 3-1 fails ka 34-1 is (-13,v3).

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