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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?

  • 0
    Hint: Where is the nth term heading to? – 2017-01-20
  • 7
    $\frac{n^2}{n}=n$ – 2017-01-20
  • 2
    And your logic is wrong : you need to find something smaller that diverges not something bigger that diverges – 2017-01-20
  • 0
    Your conclusion is right, but the comparison test does not work this way. To show the series diverges, you have to compare it to a divergent series with *smaller* terms. All you have shown is that this is smaller than a divergent sum -- but that does not mean it is divergent. – 2017-01-20
  • 0
    yes thanks everyone, silly mistake – 2017-01-20
  • 0
    $$\frac{n^2}{n+1}=(n-1)+\frac{1}{n+1}$$ and both $\sum_{n\geq 1}(n-1)$ and $\sum_{n\geq 1}\frac{1}{n+1}$ are diverging. – 2017-01-20

4 Answers 4

7

Notice that

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

7

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

4

\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.

2

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!