I think this converges only for x=0. Consider what happens when x is any small number, say 0.01
Then you have then nth term greater than 1 when
(n/3)^n * 0/01^(n-1) = (0.00333333n)^n >1
clearly that holds when 0.00333333n > 1
So, for any |x| > 0 there will come a point where the nth term exceeds 1 for sufficiently large n.
f(x) = summation from n equals 0 to infinity of the quotient of the quantity n plus 1 and 3 to the n plus 1 power times x to the nth power with an interval of convergence, –3 < x < 3. Find exactly the value of the integral from 0 to 2 of of x, dx. Your answer will be a positive integer. Type your answer in the space below (ex. 4).
2 answers
thank you so much for your help!