What is the interval of convergence of the series #sum_(n=0)^oo((-3)^n*x^n)/sqrt(n+1)#?

1 Answer
Sep 27, 2014

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]#.