Question

How do you test the series `Sigma 2/(4n-3)` from n is `[1,oo)` for convergence?

Answer 1

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)