How do you test the series #Sigma 2/(4n-3)# from n is #[1,oo)# for convergence?
1 Answer
Jan 18, 2018
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=0)^oo a_n # where#a_n = 2/(4n-3) #
So our test limit is:
# L = lim_(n rarr oo) abs( (2/(4(n+1)-3))/(2/(4n-3)) ) #
# \ \ \ = lim_(n rarr oo) abs( 2/(4n+4-3) * (4n-3)/2 ) #
# \ \ \ = lim_(n rarr oo) abs( (4n-3)/(4n+1) ) #
# \ \ \ = lim_(n rarr oo) abs( ((4n+1)-4)/(4n+1) ) #
# \ \ \ = lim_(n rarr oo) abs( 1 - 4/(4n+1) ) #
# \ \ \ = 1 #
and we can conclude that in the case the ratio test is inconclusive, (in fact the series diverges)
