Is The Series Below Convergent And If So, Find Its Limit?

For a series to be convergent there are many tests that can show it happens. Let’s use this one, known as the Ratio Test.
Let a_n = 3^n(n^2 + 1)x^n
And a_n+1 = 3^(n+1)[(n+1)^2 + 1]x^(n+1)
The premise of this test is that
lim n-> inf | [ a_n+1 / a_n ] | < 1
In words, that is "the limit as n goes to infinity of the absolute value of a_n+1 over a_n must be less than one".
What this means is that as the terms get infinitely bigger, they also get successively smaller, with each term being smaller than the preceding one.
Let's do the test (sorry if it looks so crappy in this text format):
lim n-> inf | [ (3^(n+1) ((n+1)^2 x^(n+1) ] / [ 3^n (n^2 + 1) x^n ] | < 1
lim n-> inf | 3x (n^2 + 2n + 2) / (n^2 + 1) |
Now we need to multiply the top and bottom of this fraction by n^2.
lim n-> inf | 3x (1 + 2/n + 2/n^2) / (1 + 1/n^2) | < 1
Taking the limit as n-> inf, we find that
| 3x | < 1
-1 < 3x < 1
-1/3 < x < 1/3
So what does that mean? It means that in order for this series to be convergent, and in order for the Ratio Test to be true, that the x we choose must be somewhere between negative one-third and positive one-third.
Another way to show this same result is to consider that this is a geometric series.
sum from n=0 to n = inf { 3^n (n^2 + 1) x^n }
sum from n=0 to n = inf { (3x)^n (n^2 + 1) }
Each term gets multiplied by 3x, depending on what we choose x to be. If we choose x to be anything larger or smaller than one-third, we multiply each term of the series by something larger than one, making it a divergent series.
Therefore, our assumptions and findings in the Ratio Test are true.
Finally, we need to check whether or not the endpoint of the test, x = 1/3 or x = -1/3 are valid conditions for x. How do we check? Plug them in:
x = 1/3
sum from n=0 to n = inf { (1)^n (n^2 + 1) }
sum from n=0 to n = inf { (n^2 + 1) }
That does not look like a convergent series, and in fact it is not. Therefore, x != 1/3.
How about -1/3?
x = -1/3
sum from n=0 to n = inf { (-1)^n (n^2 + 1) }
This is an alternating series because the sign changes for each successive term. To find out whether or not this converges, we need to apply the alternating series test. In order for an alternating series to converge, each successive term needs to decrease in size.
b_n = n^2 + 1
b_n+1 = n^2 + 2n + 2
lim n-> inf [ b_n+1 / b_n ]
lim n-> inf [ (n^2 + 2n + 2) / (n^2 + 1) ] => 1
Therefore, we cannot include -1/3 either, because the terms do not decrease in size.
Thus, the series is convergent only for:
-1/3 < x < 1/3

All three factors go to infinity. So the series goes to infinity. Therefore it is divergent.