How do you apply the ratio test to determine if #sum (1*3*5* * * (2n-1))/(1*4*7* * * (3n-2))# from #n=[1,oo)# is convergent to divergent?
1 Answer
the series is convergent.
Explanation:
We can apply d'Alembert's ratio test:
Suppose that;
# S=sum_(r=1)^oo a_n \ \ # , and#\ \ L=lim_(n rarr oo) |a_(n+1)/a_n| #
Then
if L < 1 then the series converges absolutely;
if L > 1 then the series is divergent;
if L = 1 or the limit fails to exist the test is inconclusive.
So our series is;
# S=sum_(n=1)^oo { 1*3*5 * ... * (2n-1) } / (1*4*7* ... * (3n-2) } #
So our test limit is:
# L = lim_(n rarr oo) | { { 1*3*5 * ... * (2n-1)* (2(n+1)-1) } / (1*4*7* ... * (3n-2)* (3(n+1)-2) } } / { { 1*3*5 * ... * (2n-1) } / (1*4*7* ... * (3n-2) } } | #
# L = lim_(n rarr oo) | { 1*3 * ... * (2n-1) (2(n+1)-1) } / { 1*4* ... * (3n-2) (3(n+1)-2) } * {1*4*7* ... * (3n-2) } / { 1*3*5 * ... * (2n-1) } | #
# L = lim_(n rarr oo) | { color(red)cancel(1*3 * ... * (2n-1)) (2(n+1)-1) } / { color(blue)cancel(1*4* ... * (3n-2)) (3(n+1)-2) } * {color(blue)cancel(1*4*7* ... * (3n-2)) } / { color(red)cancel(1*3*5 * ... * (2n-1)) } | #
# L = lim_(n rarr oo) | (2(n+1)-1) / (3(n+1)-2) | #
# \ \ \ = lim_(n rarr oo) | (2n+1)/(3n+1) | #
# \ \ \ = lim_(n rarr oo) | (2n+1)/(3n+1) *(1/n)/(1/n)| #
# \ \ \ = lim_(n rarr oo) | (2+1/n)/(3+1/n) | #
# \ \ \ = 2/3 #
And so the series is convergent.
