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Does $\sum_1^\infty \frac{n}{n^2 + 4}$ converge or diverge?

I am confused because my friend insists the series converges conditionally. I think the series diverges. Here is my process and solution:

$\sum_1^\infty \frac{n}{n^2 + 4}$

Limit Comparison Test:

Let $\sum_1^\infty \frac{n}{n^2 + 4} = \sum_1^\infty a_n$

Let $\sum_1^\infty b_n = \sum_1^\infty \frac{1}{n}$

Since $$\lim \limits_{n \to \infty} | \frac{a_n}{b_n}| = \lim \limits_{n \to \infty} | \frac{\frac{n}{n^2 + 4}}{\frac{1}{n}}| = \lim \limits_{n \to \infty} \frac{n^2}{n^2 + 4} = \lim \limits_{n \to \infty} \frac{n^2}{n^2(1 + \frac{4}{n^2})} = \lim \limits_{n \to \infty} \frac{1}{1 + \frac{4}{n^2}} = 1$$

and $1 > 0$ and $1$ is a finite number,

we can say that the behavior of $\sum_1^\infty a_n$ is the same as the behavior of $\sum_1^\infty b_n$.

Since $\sum_1^\infty b_n$ diverges ( $\sum_1^\infty \frac{1}{n}$ diverges by p-series, $p=1$), $\sum_1^\infty a_n$ diverges.

So by the Limit Comparison Test, $\sum_1^\infty \frac{n}{n^2 + 4}$ diverges.

Right?


Also, this problem was an exercise in the Ratio/Root Test section. Both tests, however seem to fail. Can the convergence be solved with the Root Test or the Ratio Test?

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    $\begingroup$ Yes, that is correct. $\endgroup$ – user28375028 Sep 16 '14 at 18:07
  • $\begingroup$ One can say that it converges improperly to $+\infty$ (and if working in the extended real line, it even converges properly to $+\infty$), and it does that unconditionally. $\endgroup$ – Daniel Fischer Sep 16 '14 at 18:08
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    $\begingroup$ Your friend doesn't know what conditional convergence means! Only a series with infinitely many strictly negative terms (and infinitely many strictly positive terms) can be conditionally convergent. $\endgroup$ – TonyK Sep 16 '14 at 18:14
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Direct comparison is a little easier here: $$\sum_{n=1}^\infty \frac{n}{n^2+4} \geq \sum_{n=2}^\infty \frac{n}{n^2+4} \geq \sum_{n=2}^\infty \frac{n}{n^2+n^2} = \frac{1}{2}\sum_{n=2}^\infty \frac{1}{n} = \infty$$

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  • $\begingroup$ So I assume we are allowed to take the sum of the series from $n = 2$ instead of $n = 1$. Also, this problem was an exercise in the Ratio/Root Test section. Both tests, however seem to fail. Can the convergence be solved with the Root Test or the Ratio Test? $\endgroup$ – user168210 Sep 16 '14 at 18:13
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    $\begingroup$ The difference between summing from 1 and from 2 is only one finite term, so the result (convergent or divergent) will not be affected. $\endgroup$ – E W H Lee Sep 16 '14 at 18:15
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    $\begingroup$ The Ratio Test is inconclusive here, because the ratio of consecutive terms tends to $1$. $\endgroup$ – TonyK Sep 16 '14 at 18:16
  • $\begingroup$ @EWHLee I see. $$Finite + Finite = Finite$$ $\endgroup$ – user168210 Sep 16 '14 at 18:16
  • $\begingroup$ @gragas: Yes. But now I updated it so that the argument is more complete. $\endgroup$ – E W H Lee Sep 16 '14 at 21:32
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Write $\frac{n}{n^2 + 4} = \frac1{n + \frac{4}{n}}$. Note that $ n + \frac{4}{n} \le n + 4 $. Use comparison test.

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Since your $a_n \geqslant \frac {1} {2n}$ except at $n = 1$ and $\sum \frac {1} {2n}$ diverges, you establish that $\sum a_n$ diverges.

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