how do you work out if this is a convergent series?

Question

∞
∑ 4n^2 + 2n -1/ 2n^2 - 1
n=1

Answer 1

lim [ ( 4n^2 + 2n - 1 ) / ( 2n^2 - 1 ) ] =
n→∞
lim [ ( 8n + 2 ) / ( 4n - 1 ) ] = 4/2 = 2
n→∞

The series diverges.

Answer 2

Why don´t you add parentheses to make clear what´s the denominator and what the numerator?? As you wrote it looks like ONLY 2n^2 is the denominator of -1 and all the rest is given without fractions, and I´m sure you didn´t mean this.
Tonio

Answer 3

It looks like the value of each individual term tends to 2. For example, the 1st term is 5, 10th term is 2.105528, 100th term is 2.010051, and the 1000th term is 2.001001, in other words, closer and closer to 2.

This is not a proof, but it does look like a divergent series, because in a convergent series, the values of the individual terms would need to tend to 0, although that is not the only criteria - it is still possible for a series to be divergent even if its terms do tend to 0 but don't tend quickly enough.