Why the following series
$$\sum_{n=1}^{\infty} \frac{\ln n }{\sqrt{n^3-n+1}}$$
converges?
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Note that by L'Hospital's Rule, or other methods,$\lim_{n\to \infty} \frac{\log n}{n^{1/4}}=0$. So if $n$ is large enough, $\log n \lt n^{1/4}$.
Note also that $n^3-n+1\gt n^3/4$. Thus if $n$ is large enough, then $$\frac{\log n}{\sqrt{n^3-n+1}}\lt \frac{2}{n^{5/4}}.$$ It follows by Comparison that our series converges.
André Nicolas
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A related problem. Note that, you can compare as
$$ \frac{\ln n }{\sqrt{n^3-n+1}} \sim \frac{\ln n }{\sqrt{n^3}},$$
and the series
$$ \sum_{n}\frac{\ln n }{\sqrt{n^3}} $$
converges by integral test.
Mhenni Benghorbal
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Why is $1/\sqrt{n^3 - n +1} \approx 1/\sqrt{n^3}$? – Lemon Jun 20 '13 at 01:15
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1@sidht $$\frac{1}{\sqrt{n^3-n+1}} =\frac{1}{\sqrt{n^3(1-1/n^2 +1/n^3)}} $$ – Shuhao Cao Jun 20 '13 at 01:22
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@ShuhaoCao: Thanks for answering him. – Mhenni Benghorbal Jun 20 '13 at 06:18