Question

Find the interval of convergence for the power series. Do a check for the endpoints. "(x+2) 0 2"

Answer 1

Fact: Ratio test] : (statement) Let San with lim anh ant exist be a series bee I ho an then 1. If < < 1 the series is convers ent 2. If L> 1 the series is divergent test in conclusive. IS 3. If L= CS #camied with lamScanner

Let Een n² (x+2) an o then an n² (8+2) 2n Consideü lim anti noo an (n+02 (x+2)1+ lim n 2 n2(x+2jh ht hoo lim nale hooo = **+2+ lim ft+toy hooo 1 x+2) श By ratio test ous series an D=0 is convergent if lim lant nooo an <1 or 1 X+21 < 1 Te 1X +21 < 2 ie 26X+242 i e -4 < X<o Scanned with CS Bandera CamScanner

Discussion at end point! at X=-4 eg. S an = n² (-2)" put x = -4 X = - 4 in (L we get an h-o ( -n2 hao an Eighn 2. Scanned CamScanner: n=0 then hoo ht an no lim an lim corn? does not exist hooo by our series So nth term test :is conteagent divergent So at X=-4 n-(x+2) is divergent an o us power senes Hat put x=0 in eq. we get n2 a h2o gh 2 n² a

then an = n2 lim ana lim n' = (70) neo ht So by nith term test is diversent our series- n²(x+2) So al x=0 our series power on 5-6 is divergent Conclusion: Internal of Convergence is. (-407 ES Scanned WitM CamScanner