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In lecture, we were asked:

Does $ \sum \limits_{n=0}^\infty \frac{ 2n^3+ 3n -8 }{ n^5-5n^3-n^2+2 }$ converge?

$a_n = \frac{ 2n^3+ 3n -8 }{ n^5-5n^3-n^2+2 }$

In discussing strategies for applying a comparison test for $a_n$ we rejected several options for choosing a bound for $b_n$. The strategy is to choose an expression larger in the numerator and smaller in the denominator for $b_n$ If I understand correctly, using $2n^5$ in the denominator for $b_n$ doesn't work, because, as n gets large, we have

$$ \frac{2n^3}{n^5} \not < \frac{3n^3}{2n^5}$$

$$ 2 \not < \frac{3}{2} $$

If this is the case, couldn't we choose $b_n = \frac{10n^3}{2n^5}$?

Plugging the first inequality into Wolfram Alpha was not enlightening. enter image description here

From the lecture:

To apply the comparison test, we need to bound the numerator from above and the denominator from below.

For the numerator of $b_n$, $2n^3$ won't work because we have $3n$ in the numerator of $a_n$. $3n^3$ is large enough.

$2n^5$ in the denominator of $b_n$ won't work.

$a_n = \frac{ 2n^3+ 3n -8 }{ n^5-5n^3-n^2+2 } \ \substack{ <\\ \\ >} \ \frac{3n^3}{2n^5} =b_n$

Using $n^4$ in the denominator won't work because this gives us a harmonic series for $b_n$, which is divergent.

$a_n = \frac{ 2n^3+ 3n -8 }{ n^5-5n^3-n^2+2 } \ \substack{ <\\ \\ >} \ \frac{3n^3}{n^4}=b_n $

Using asymptotic analysis, we get $a(x)= \frac{2}{x^2} + \mathcal{O}(\frac{1}{x^4})$, which gives us an integral with p=2, which converges.

maogenc
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2 Answers2

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The easiest way to deal with this situation is to use the Limit Comparison Test, since then you don't have to worry about choosing appropriate constants. However, if you want to use the Comparison Test directly, trying a comparison with $\sum_{n=1}^\infty \frac{a}{n^2}$ and solving $\frac{2n^3+3n-8}{n^5-5n^3-n^2+2}\le \frac{a}{n^2}$ by cross-multiplying will show that any choice of $a$ with $a>2$ will work, since $2n^5+3n^3-8n^2\le an^5-5an^3-an^2+2a$ for n sufficiently large if $a>2$.

user84413
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We do an explicit comparison, since that's what the question seems to ask for, but there are simpler ways.

For $n\gt 1$, the top is positive and $\lt 2n^3+3n^3=5n^3$.

The bottom is trickier. Rewrite it as $$\frac{1}{2}n^5+ \frac{1}{2}n^5-5n^3-n^2+2.$$

Note that if $n\gt 10$, then $\frac{1}{2}n^5-5n^3-n^2+2\gt 0$. This is because then $\frac{1}{2}n^5 \gt 50n^3$. (We could get away with something cheaper than $10$, but why bother?). So the bottom, when $n\gt 10$, is $\gt \frac{1}{2}n^5$.

What we did here was to despatch half of $n^5$ to go crush $5n^3+n^2$. After they are crushed, the bottom is bigger than the remaining $\frac{1}{2}n^5$.

So for $n\gt 10$ we have that our expression is positive and $\lt \frac{5n^3}{x^5/2}=\frac{10}{n^2}$. Good enough to show convergence.

André Nicolas
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