We can apply the ratio test stating that a necessary condition for the series:
sum_(n=1)^oo a_n
to converge is that:
L = lim_(n->oo) abs(a_(n+1)/a_n) <= 1
If L<1 the condition is also sufficient and the series converges absolutely.
Lat's calculate the ratio for the series:
abs(a_(n+1)/a_n) = abs ( frac ((((n+1)+1)(x+4)^(n+1))/(7^(n+1)(5(n+1)-3))) (((n+1)(x+4)^n)/(7^n(5n-3)))
abs(a_(n+1)/a_n) = ((n+2)abs(x+4)^(n+1))/(7^(n+1)(5n+2)) (7^n(5n-3))/((n+1)abs(x+4)^n)
abs(a_(n+1)/a_n) = (n+2)/(n+1) abs(x+4)^(n+1) / abs(x+4)^n 7^n/7^(n+1) (5n-3)/(5n+2)
abs(a_(n+1)/a_n) = 1/7abs(x+4) ( (n+2)/(n+1))((5n-3)/(5n+2))
Passing to the limit:
lim_(n->oo) abs(a_(n+1)/a_n) =1/7abs(x+4)
Thus we have that the series is absolutely convergent for:
1/7abs(x+4) < 1
abs(x+4) < 7
-11 < x < 3
and divergent for x<-11 and x>3.
For abs(x+4) = 7 the test is inconclusive and we have to analyze case by case:
(1) x= -11
a_n= ((n+1)(-11+4)^n)/(7^n(5n-3)) = (-1)^n (n+1)/(5n-3)
(2) x= 3
a_n= ((n+1)(3+4)^n)/(7^n(5n-3)) = (n+1)/(5n-3)
In both cases we have:
lim_(n->oo) a_n != 0
so the series doesn't satisfy Cauchy's necessary condition and cannot be convergent.
In conclusion the series is convergent for x in (-11,3)
