prove that the series $\sum \frac {n^2}{n+1}$ diverges.

Question

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?

Answer 1

Notice that

$$ \lim_{n \to \infty} \frac{n^2}{n+1} \;\; =\;\; \lim_{n \to \infty} \frac{n}{1 + \frac{1}{n}} \;\; =\;\; \left ( \lim_{n \to \infty} \frac{1}{1 + \frac{1}{n}} \right ) \left ( \lim_{n \to \infty} n \right ) \;\; =\;\; \lim_{n \to \infty} n. $$

Answer 2

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 $\lim_{n\to\infty}\frac{n^2}{n+1}=+\infty$.

Answer 3

\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.

Answer 4

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!