Testing A Series For Convergence Determine whether the series 
$\sum_{n=0}^{\infty} \frac{3n^2 + 2n + 1}{n^3 + 1}$ with n from 0 to infinity 
converges or diverges.
So far I thought about dividing the numerator by the denominator, but that got very messy.
I thought about comparing that to the series of $\frac{1}{k^3 + 1}  but then I got stuck.
Also, a related question. 
A theorem states that if the limit of a series as n approaches infinity is not equal to zero, the series diverges. However it states that the series is from n=1 to infinity. Would it also apply in this case where it goes from n=0 to infinity?
Thanks!
 A: The critical question is whether or not $\sum 3n^2/(n^3+1)$ converges.
The other terms, $\sum 2n/(n^3+1)$ and $\sum 1/(n^3 + 1)$ will definitely converge, so the real question is just that first term.
I will claim that the first term diverges because for each term where n > 0,
$\sum 3n^2/(n^3+1)$ is greater than the corresponding term of $\sum 3n^2/(3n^3)$
which matches each term of $\sum 1/n$.
Compare with harmonic series...
For the second part of your question, as Trogdor mentioned, yes the same theorem applies, but it is true not for the sequence approaching a value, but the sequence of terms (or summands as Trogdor suggests).
Trivial example: $\sum 1$  The sequence of terms clearly approaches the value 1, which is not 0 (in fact, it actually equals 1 for each term).  As you can probably tell, this sequence diverges.
In general, if $\sum_{n=1}^\infty \Phi (n)$ diverges and $\Phi(0)$ is defined, then clearly $\Phi (0) + \sum_{n=1}^\infty \Phi(n)$ diverges, but this last result is just $\sum_{n=0}^\infty \Phi(n)$, so the theorem applies here as well.
A: The $n$-th term is $\ge \frac{3n^2}{2n^3}$, which is $\frac{3}{2}\cdot\frac{1}{n}$. Now compare with the harmonic series to deduce divergence. 
As to your second question, the convergence/divergence of a series is not affected if we alter a finite number of terms. So whether we sum from $0$ to $\infty$ or from $1$ to $\infty$ makes no difference. 
