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].
