The interval of convergence is #(-1/3,1/3]#.
Let #a_n={(-3)^nx^n}/{sqrt{n+1}}#. #Rightarrow a_{n+1}={(-3)^{n+1}x^{n+1}}/{sqrt{n+2}}#.
By Ratio Test,
#lim_{n to infty}|{a_{n+1}}/{a_n}|
=lim_{n to infty}|{(-3)^{n+1}x^{n+1}}/{sqrt{n+2}}cdot{sqrt{n+1}}/{(-3)^nx^n}|#
by cancelling out common factors,
#=lim_{n to infty}|{-3x sqrt{n+1}}/{sqrt{n+2}}|#
by pulling #|-3x|=3|x|# out of the limit,
#=3|x| lim_{n to infty}\sqrt{{n+1}/{n+2}}#
by dividing the numerator and the denominator by #n#,
#=3|x| lim_{n to infty}\sqrt{{1+1/n}/{1+2/n}}=3|x|sqrt{{1+0}/{1+0}}=3|x|<1#
By dividing by 3,
#Rightarrow |x|<1/3 Leftrightarrow -1/3 < x < 1/3#
So, the power series converges at least on #(-1/3,1/3)#.
Now, we need to check the endpoints.
When #x=-1/3#, the series becomes
#sum_{n=0}^infty{(-3)^n(-1/3)^n}/{sqrt{n+1}}=sum_{n=0}^infty1/{sqrt{n+1}}=sum_{n=1}^infty1/n^{1/2}#,
which is a divergence p-series since #p=1/2 le 1#
So, #x=-1/3# should be excluded.
When #x=1/3#, the serie becomes
#sum_{n=0}^infty{(-3)^n(1/3)^n}/{sqrt{n+1}}=sum_{n=0}^infty(-1)^n/{sqrt{n+1}}#,
which is a convergent alternating series by Alternating Series Test.
#1/sqrt{n+1} ge 1/sqrt{n+2}# and #lim_{n to infty}1/sqrt{n+1}=0#
So, #x=1/3# should be included.
Hence, the interval of convergence is #(-1/3,1/3]#.
