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I am stuck on this problem. I have checked on wolfram alpha, and the calculator says that this series is divergent. However, I can't find a way to prove it.

The problem:

$$\sum\limits_{n=1}^{\infty} \frac{n^2-1}{n^3+1}$$

So I tried doing this problem with the direct comparison test.

I have concluded that

$$ \frac{n^2}{n^3} >\frac{n^2-1}{n^3+1} $$ $$ b_{n}> a_{n} $$

$b_{n}$ is a p-series, where p=1. This means that the series $b_{n}$ is divergent, making the test direct comparison test inconclusive. $$ \frac{n^2}{n^3} = \frac{1}{n} $$

I appreciate any help... thanks! :)

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    $\begingroup$ It's divergent. $\endgroup$
    – Botnakov N.
    Jan 10, 2021 at 23:12
  • $\begingroup$ @ Botnakov N. - that’s the best comeback I’ve heard all week. $\endgroup$
    – Adam Rubinson
    Jan 10, 2021 at 23:14
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    $\begingroup$ Show that for large $n$ you have $a_n \ge c {1 \over n}$ for some constant $c>0$. $\endgroup$
    – copper.hat
    Jan 10, 2021 at 23:21
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    $\begingroup$ The series $\sum_n {1 \over n}$ is divergent, so by comparison we have $\sum_{n \ge N} a_n \ge c \sum_{n \ge N} {1 \over n}$. $\endgroup$
    – copper.hat
    Jan 10, 2021 at 23:29
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    $\begingroup$ Another way is to write $$ \frac{n^2-1}{n^3+1}=\frac{n-1}{n^2-n+1}=\frac{1}{n+1/(n-1)}\ge\frac{1}{n+1} $$ or you can do the whole "add and subtract $1/n$" trick. $\endgroup$
    – anon
    Jan 10, 2021 at 23:38

2 Answers 2

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Note that

$$\frac{n^2-1}{n^3+1} \geq \frac{n^2 - \frac{n^2}{2}}{n^3+n^3} = \frac{1}{4}\frac{n^2}{n^3} =\frac{1}{4n}. $$

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  • $\begingroup$ I'm trying my best to understand. Why did you add $\frac{-n^2}{2} $ to the numerator and $n^3$ to the denominator? $\endgroup$
    – galaxyworks
    Jan 10, 2021 at 23:53
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    $\begingroup$ If you reduce the numerator, you get something smaller. If you increase the denominator, you get something smaller. If you do both, you get something smaller. $\endgroup$
    – saulspatz
    Jan 11, 2021 at 0:00
  • $\begingroup$ I understand. But in which way can that help us prove that the original sequence is divergent? @saulspatz $\endgroup$
    – galaxyworks
    Jan 11, 2021 at 0:05
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    $\begingroup$ What's 1/4 of infinity? $\endgroup$
    – B. Goddard
    Jan 11, 2021 at 0:18
  • $\begingroup$ 1/4 is insignificant since infinity is an infinitely large number? $\endgroup$
    – galaxyworks
    Jan 11, 2021 at 0:23
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$$ \frac{n^2-1}{n^3+1} = \frac{n^2+1 -2}{n^3+1} = \frac{n^2+1}{n^3+1} -\frac{2}{n^3+1} \geq \frac{1}{n} - \frac{2}{n^3+1} \quad \forall\ n \in \mathbb{N}. $$

Therefore,

\begin{align}\sum\limits_{n=1}^{\infty} \frac{n^2-1}{n^3+1} = \sum\limits_{n=1}^{\infty}\left(\frac{n^2+1}{n^3+1} -\frac{2}{n^3+1}\right)\\ \\ = \sum\limits_{n=1}^{\infty}\frac{n^2+1}{n^3+1}\ -\ \sum\limits_{n=1}^{\infty}\frac{2}{n^3+1} \\ \\ \geq \left(\sum\limits_{n=1}^{\infty} \frac{1}{n}\right)\ -\ L \\ \end{align}

for some $L\in \mathbb{R}.$

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