prove that the series $\sum \frac {n^2}{n+1}$ diverges. Looking for some help with the following problem,
I need to prove that the series $\sum \frac {n^2}{n+1}$ diverges.
My solution was:
I decided to use the comparison test and noticed that,
$\frac{n^2}{n+1}<\frac {n^2}{n}=\frac {1}{n}$ and $\sum \frac{1}{n}$ diverges therefore $\sum \frac {n^2}{n+1}$ diverges
Is this a correct solution or do I need to go about it in a different way?
 A: Notice that 
$$
\lim_{n \to \infty} \frac{n^2}{n+1} \;\; =\;\; \lim_{n \to \infty} \frac{n}{1 + \frac{1}{n}} \;\; =\;\; \left ( \lim_{n \to \infty} \frac{1}{1 + \frac{1}{n}} \right ) \left ( \lim_{n \to \infty} n \right ) \;\; =\;\; \lim_{n \to \infty} n.
$$
A: First, if $\sum a_n$ and $\sum b_n$ are two series with positive terms such that $a_n\leq b_n$ and $\sum b_n$ diverges, then there's nothing we can say about the convergence of $\sum a_n$. We need more information.
Second, if a series $\sum a_n$ converges, then one must have $\lim_{n\to\infty}a_n=0$.
Hence the series is divergent since $\lim_{n\to\infty}\frac{n^2}{n+1}=+\infty$.
A: \begin{align}
S(N) &= \sum_{n=1}^N \frac{n^2}{n+1}\\
&>\sum_{n=1}^N \frac{n^2}{n+n}\\
&=\frac{1}{2}\sum_{n=1}^N n\\
&=\frac{1}{2}\cdot \frac{N(N+1)}{2}
\end{align}
Since the sequence of partial sums is unbounded, the series diverges.
A: Check the degree of the numerator vs. the degree of the denominator. Since $n^2$ (of degree 2) has a higher degree than $n+1$ (of degree 1), the series is divergent.
The reasoning here is that $n^2$ grows much faster than $n+1$, therefore, when dividing the two when both evaluated at a very large number, the higher order term tends to overpower the other one.
Just a little intuition. Hope this helps!
