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I came across the following series on a test today: $$\sum_{n=1}^{\infty}\frac {(n^3-n^2+3)(n^{5/3})} {n^5+10^{10}n^4-1}$$ The question was to figure out whether this converged absolutely, conditionally, or diverged.

Well, to start off I tried distributing the $(n^{5/3})$ and then rewriting the series with the exponents in fractional form. So it was $$\sum_{n=1}^{\infty}\frac {(n^{14/3}-n^{11/3}+3n^{5/3})} {n^{15/3}+10^{10}n^{12/3}-1}$$

I figured that it probably diverges because it "looks like" the p-series $\sum_{n=1}^{\infty}\frac 1 {n^{1/3}}$ but I didn't know how to test it further and actually show that it diverges. So after that I was stuck!

Can someone please help me understand how to do this problem?

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3 Answers 3

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Hint. You may just write, as $n \to +\infty$, $$ \frac {(n^3-n^2+3)(n^{5/3})} {n^5+10^{10}n^4-1} \sim \frac {n^3(n^{5/3})} {n^5}=\frac{1}{n^{1/3}} $$ then your initial series is divergent.

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  • $\begingroup$ I liked both your and learnmore's answers, but I slightly preferred this one because of its simplicity. $\endgroup$
    – Asker
    Commented Feb 4, 2015 at 4:20
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Compare the series with $v_n=\dfrac{1}{n^\frac{1}{3}}$ and check the limit $\dfrac{u_n}{v_n}$ as $n \rightarrow \infty$ .You will find it to be 1 which is non-zero and finite.Hence the two series converge or diverge together.But by $p-$ test $v_n$ diverges

Thus original series $u_n$ also ...

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You are correct. Divide the numerator and denominator by $n^{14/3}$ and you get:

$$\sum_n \frac{1-n^{-1}+3n^{-3}}{n^{1/3}+10^{10}n^{-2/3}-n^{-14/3}}$$

When $n^{2/3}>10^{10}$, the numerator is bigger than $1/2$, and the denominator is less than $2n^{1/3}$. So for large enough $n$, the terms are bigger than $\frac{1}{2n^{1/3}}$.

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