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To test the convergence of the series $\sum_{n=2}^\infty\frac{1}{n^p-n^q}$ I tried the limit comparison test. $0<q<p.$ My $a_n=n^p-n^q$ and my $b_n=\frac{1}{n^p-n^q}$. The limit comparison test says:

If $$\lim_{n\to\infty}\frac{b_n}{a_n}=1$$ then $\sum a_n$ converges iff $\sum b_n$ converges.

Now, since the $\lim_{n\to\infty}a_n=\infty \ne0$, $\sum a_n$ can not converge. And if $\sum a_n$ does not converge then $\sum b_n$ can not converge. After my try I discovered that I'm wrong according to the solutions. Why and where am I wrong?

Thank you.

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  • $\begingroup$ @Andre Can you be more explicit please? $\endgroup$ – Charlie Jan 25 '14 at 16:02
  • $\begingroup$ Wrote an answer. $\endgroup$ – André Nicolas Jan 25 '14 at 16:03
  • $\begingroup$ Idea to use the limit comparison test: the dominant (greatest when $n$ grows) term in $n^p-n^q$ is...? $\endgroup$ – Martín-Blas Pérez Pinilla Jan 25 '14 at 16:10
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    $\begingroup$ Yes. Now, do $\lim((1/(n^p-n^q))/(1/n^p))$. $\endgroup$ – Martín-Blas Pérez Pinilla Jan 25 '14 at 16:19
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    $\begingroup$ What are your conditions on $p$ and $q$? Clearly, if $p>2$, the series converges since we can bound $\frac{1}{n^p-n^q} \leq \frac{1}{n^p/2} \leq \frac{1}{n^{p-1}}$. Also, if $p\leq1$, the series is divergent, since $\frac{1}{n^p-n^q} \geq \frac{1}{n^p}$. So it remains to check convergence for $1<p\leq 2$ by using the limit comparison test and $a_{n} = n^{-p}$ and $b_{n} = \frac{1}{n^p-n^q}$. $\endgroup$ – Chris K Jan 25 '14 at 19:00
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Hint: Wrong $a_n$. It should be $\frac{1}{n^p}$. Then use standard facts about $p$-series.

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  • $\begingroup$ Ok, could you explain me please why the $$\lim_{n\to\infty}\frac {n^p}{n^p-n^q}=1?$$ Thank you... $\endgroup$ – Charlie Jan 25 '14 at 16:18
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    $\begingroup$ Divide top and bottom by $n^p$. We get $\frac{1}{1+\frac{1}{n^{p-q}}}$. The $\frac{1}{n^{p-q}}$ part goes to $0$. $\endgroup$ – André Nicolas Jan 25 '14 at 16:21
  • $\begingroup$ My last question is: why my $a_n$ did not work? $\endgroup$ – Charlie Jan 25 '14 at 16:35
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    $\begingroup$ Most $a_n$ won't work, the limit of $\frac{b_n}{a_n}$ for your choice of $a_n$ is $0$, not $1$. We want to choose an $a_n$ that behaves in the long run like $b_n$, in the sense that the ratio has a non-zero limit. $\endgroup$ – André Nicolas Jan 25 '14 at 16:41
  • $\begingroup$ Are you saying that the $\lim_{n\to\infty}\frac{n^p-n^q}{n^p-n^q}=0?$ If so why? Isn't that ratio, and its limit as $n\to\infty$, equal to 1? $\endgroup$ – Charlie Jan 25 '14 at 17:35

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