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Comparison test $\sum_{n=1}^{\infty} \frac{1}{2n+7}$

My textbook did the limit comparison test to do this but cant you just do this using direct comparison test?

What I did

For $n \geq 1, \forall n \in \mathbb N$, $a_n = \frac{1}{2n+7} \leq \frac{1}{n} = b_n$

Since $$0 < a_n \leq b_n, \forall n \in \mathbb N$$

Direct comparison test applies

Consider $\sum_{n=1}^{\infty} \frac{1}{n}$. Since this is a p-series with magnitude one we know that $\sum b_n$ diverges. Therefore by the direct comparison test $\sum a_n$ diverges as well.

Usually my textbook does the easiest way so I don't know.

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  • $\begingroup$ What did your textbook do? $\endgroup$ – DHMO Apr 13 '17 at 7:36
  • $\begingroup$ Same $b_n$ but they used limit comparison to show that $\lim_{n\to\infty} \frac{a_n}{b_n}$ = # $\in (0, \infty)$. Why is this neccesary? $\endgroup$ – user349557 Apr 13 '17 at 7:36
  • $\begingroup$ Arguably both are of the same difficulty, though the limit comparison test has wider application -- the idea of considering how a function/sequence behaves in the "long run" is important. Here, the intuition is to realize $1/(2n+7)$ basically grows like $1/n$ -- no need or any bounding. $\endgroup$ – user217285 Apr 13 '17 at 7:38
  • $\begingroup$ So both solutions are correct? $\endgroup$ – user349557 Apr 13 '17 at 7:39
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Yes, you can use direct Comparison Test.

For all $n\geq 1$, we get $$\frac{1}{2n+\color\red 7}\geq\frac{1}{2n+\color\red{7n}}=\frac{1}{9}\cdot\frac{1}{n}.$$ Since the series $$\sum_{n=1}^{\infty}\bigg(\frac{1}{9}\cdot\frac{1}{n}\bigg)$$ is a divergent series, it follows from the Comparison Test that the series $$\sum_{n=1}^{\infty}\frac{1}{2n+7}$$ is also divergent.

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  • $\begingroup$ Thx I really like this way $\endgroup$ – user349557 Apr 13 '17 at 7:47
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If you have got $b_{n} \leq a_{n}$, then $\sum_{n}b_{n}$ divergent implies $\sum_{n}a_{n}$ diverges. But you have got $b_{n} \geq a_{n}$; so there is no guarantee in this case. To apply comparison test, you have to seek another sequence $c_{n}$ ensuring that $a_{n} \geq c_{n}$ and that $\sum_{n}c_{n}$ diverges.

However, the limit comparison test can get you what you want immediately. Recall that the theorem states that if $a_{n},b_{n} \geq 0$ for all integers $n > 0$ and if $\lim_{n}a_{n}/b_{n} > 0$ then either $\sum_{n}a_{n}$ and $\sum_{n}b_{n}$ both converges or both diverges. Note that $$ \frac{1}{2n+7} \times n \to \frac{1}{2} > 0. $$

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  • $\begingroup$ Oh so it doesnt work the same way as convergence? I get it, thank you. $\endgroup$ – user349557 Apr 13 '17 at 7:46
  • $\begingroup$ Yeah by looking at a proof of comparison test you will see immediately. If instead $\sum_{n}b_{n}$ converges then $a_{n} \geq b_{n}$ implies $\sum_{n}a_{n}$ converges. $\endgroup$ – Benicio Apr 13 '17 at 7:47

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