Vtyjkyuk

For what values of $x$ in the following series, does the series converge?

beginalignsum^infty_n=1dfrac(-1)^n x^nn[log (n+1)]^2,;;-3<x<17 endalign

MY TRIAL

beginalignlimlimits_nto inftyleft|dfrac(-1)^n+1 x^n+1(n+1)[log (n+2)]^2cdotdfracn[log (n+1)]^2(-1)^n x^nright|&=|x|limlimits_nto inftyleft|dfracnn+1cdotleft[dfraclog (n+1)log (n+2)right]^2right|\&=|x|limlimits_nto inftyleft(dfracnn+1right)cdotlimlimits_nto inftyleft[dfraclog (n+1)log (n+2)right]^2\&=|x|limlimits_nto inftyleft(dfracnn+1right)cdotleft[limlimits_nto inftydfraclog (n+1)log (n+2)right]^2\&=|x|left[limlimits_nto inftydfrac1n+1cdot n+2right]^2\&=|x|endalign
Hence, the series converges absolutely for $|x|<1$ and diverges when $|x|>1$.

When $x=1,$
beginalignsum^infty_n=1dfrac(-1)^n n[log (n+1)]^2<infty;;textBy Alternating series testendalign
When $x=-1,$
beginalignsum^infty_n=1dfrac1n[log (n+1)]^2<infty;;textBy Direct comparison testendalign
Hence, the values of $x$ for which the series converges, is $-1leq xleq 1.$

I'm I right? Constructive criticisms will be highly welcome! Thanks!

You are correct. This is a "variation on the theme".

For $|x|>1$
$$lim_nto +inftydfrac^nn[log (n+1)]^2=+infty$$
and the series is divergent.

For $|x|leq 1$, by direct comparison, the series is absolutely convergent
$$sum^infty_n=1dfrac^nn[log (n+1)]^2leq sum^infty_n=1dfrac1n[log (n+1)]^2<infty.$$

Yes it is correct, for the limit from here we can proceed as follow

$$ldots=|x|limlimits_nto inftyleft|dfracnn+1cdotleft[dfraclog (n+1)log (n+2)right]^2right|
=|x|limlimits_nto inftydfrac11+1/ncdotleft[dfraclog n+log (1+1/n)log n+log (1+2/n)right]^2=|x|cdot1=|x|$$