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$$\sum_{n=1}^\infty \left(\frac{n^4}{n^5+7}\right)$$

I try this series by the comparison test with $a_n\le b_n,$ $a_n=\frac{n^4}{n^5+7}$ and $b_n=\frac{n^4}{n^5}=\frac{1}{n}$

then $b_n$ diverges, dose the series diverges ?

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  • $\begingroup$ To prove that $\sum a_n$ diverges, you'd need $b_n\leq a_n$ and $\sum b_n$ diverges. You can't use $b_n=\frac{1}{n}$, but you can use a related sequence. $\endgroup$ Feb 13, 2018 at 16:15
  • $\begingroup$ You can use the limit comparison test. $\endgroup$
    – saulspatz
    Feb 13, 2018 at 16:16
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    $\begingroup$ I use the limit comparison test then the series diverges $\endgroup$
    – Beginner
    Feb 13, 2018 at 16:24
  • $\begingroup$ @Beginer Yes the limit comparison test is the right way! $\endgroup$
    – user
    Feb 13, 2018 at 16:32

4 Answers 4

4
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A bit more formally:

Let $n \ge 2:$

$a_n=\dfrac{n^4}{n^5+7} \gt \dfrac{n^4}{n^5+n^5} =\dfrac{1}{2n}.$

Since $(1/2) \sum \dfrac{1}{n}$ diverges,

$\sum \dfrac{n^4}{n^5+7}$ diverges (comparison test).

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$$ \frac{n^4}{n^5+7}\underset{(+\infty)}{\sim}\frac{1}{n} $$

What can you say about $\displaystyle \sum_{n \geq 1}^{ }\frac{1}{n}$?

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  • $\begingroup$ the harmonic series. $\endgroup$
    – Beginner
    Feb 13, 2018 at 16:21
  • $\begingroup$ Yes, and does it converge or not ? Because your series will have the same behavior as the harmonic series. $\endgroup$
    – Atmos
    Feb 13, 2018 at 16:22
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    $\begingroup$ The series diverges. $\endgroup$
    – Beginner
    Feb 13, 2018 at 16:25
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Hint: $$\frac{n^4}{n^5+7}\geq\frac{n^4}{8n^5}=\frac{1}{8n}$$ for all $n\geq1$.

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As an alternative to the limit comparison test note that by binomial series

$$\frac{n^4}{n^5+7}=\frac{1}{n+\frac7{n^4}}=\frac1n\left(1+\frac7{n^5}\right)^{-1}\ge \frac1n\left(1-\frac7{n^5}\right)=\frac1n-\frac7{n^6}$$

thus $\sum_{n=1}^{\infty} \left(\frac{n^4}{n^5+7}\right)$ diverges.

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