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I have series $\sum_{n=2}^{\infty} \frac{n+1}{n^3-1} $

I think that comparison test would give the easiest solution but i'm not sure how to apply it. I know that i need to find larger sum to prove convergence or smaller to prove divergence, so

$\frac{1}{n} $ would give us larger sum (correct me if i am wrong), and we know that $\sum_{n=0}^{\infty} \frac{1}{n} $ diverges, therefore our series from beginning of question diverges, but wolfram alpha said that $\sum_{n=2}^{\infty} \frac{n+1}{n^3-1} $ by comparison test converges.

Where am i making mistake, and what is the best way to apply comparison test on $\sum_{n=2}^{\infty} \frac{n+1}{n^3-1} $?

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  • $\begingroup$ try $\sum_{n=2}^{\infty} \frac{2}{n^2+n+1}$. $\endgroup$ – Yiyi Rao Nov 10 '16 at 17:48
  • $\begingroup$ If you have $0\le u_n\le v_n$, and $\sum v_n$ is not convergent, then you can't say anything about $\sum u_n$. $\endgroup$ – Nicolas FRANCOIS Nov 10 '16 at 17:52
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    $\begingroup$ $\frac{n+1}{n^3-1}\sim\frac{1}{n^2}$, so $\sum u_n$ is convergent. $\endgroup$ – Nicolas FRANCOIS Nov 10 '16 at 17:54
  • $\begingroup$ The comparison test says if you're "bigger" than a divergent, then you're divergent, and if you're "smaller" than a convergent, then you're convergent. Remember that for non-negative series, convergence and divergence are basically "size" concepts. $\endgroup$ – AJY Nov 10 '16 at 17:56
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    $\begingroup$ @TarsNolan Yes, but a constant factor catches that. If you compare to $\frac{2}{n^2}$, you have $\frac{n+1}{n^3-1} < \frac{2}{n^2}$ for $n \geqslant 2$. If you pick any $c > 1$, you have $\frac{n+1}{n^3-1} < \frac{c}{n^2}$ for all large enough $n$. $\endgroup$ – Daniel Fischer Nov 10 '16 at 18:26
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Compare the given series with $\sum_{n=1}^\infty v_n=\sum\dfrac{1}{n^2}$.

If $u_n=\dfrac{n+1}{n^3+1}$ and $v_n=\dfrac{1}{n^2}$

then $\dfrac{u_n}{v_n}=\dfrac{(n^2)( n+1)}{n^3+1}=\dfrac{1+\dfrac{1}{n}}{1+\dfrac{1}{n^3}}\to 1<\infty$ as $n\to \infty$

Since $\sum_{n=1}^\infty \dfrac{1}{n^2}$ converges so does $\sum u_n=\sum\dfrac{n+1}{n^3+1}$

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What about exploiting creative telescoping? For any $n\geq 2$ we have $$ 0\leq \frac{n+1}{n^3-1} \leq \frac{1}{n(n-1)}=\frac{1}{n-1}-\frac{1}{n}\tag{1}$$ hence: $$ 0\leq \sum_{n=2}^{N}\frac{n+1}{n^3-1}\leq \sum_{n=2}^{N}\frac{1}{n(n-1)} = 1-\frac{1}{N}<1.\tag{2}$$

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