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

Question

Cameron Alexander · Accepted Answer

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}}#. Using the 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 eliminating common elements, #=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 three, #Rightarrow |x|<1/3 Leftrightarrow -1/3 < x < 1/3# So, the power series converges at least on #(-1/3,1/3)#. We must now examine 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, according to the Alternating Series Test, is a convergent alternating series. #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]#.

Isabella Ackerman · Answer

undefined The interval of convergence of the series is (-1/3, 1/3).