1
$\begingroup$

$$\sum_{n=2}^\infty \frac1{n\ln n}$$ diverges by the integral test and p-series.

But,

$$\sum_{n=2}^\infty \frac1{n^{1.1}}$$ converges by p-series, and is greater than the first series, right? So, by the direct comparison test for series, the first series should converge.

I know that a divergent series can't be smaller than a convergent series, so this must be wrong somehow. I think that

$$\frac1{n\ln n} < \frac1{n^{1.1}}$$

must be wrong, but I don't know how to show it. From what I can tell, the first series is larger at first, but at a very large n, is much smaller. Thanks for the help!!

$\endgroup$
1
  • 3
    $\begingroup$ You're wrong: $n\log n=o\bigl(n^{1.1}\bigr)$, so $\dfrac1{n^{1.1}}=o\Bigl(\dfrac1{n\log n}\Bigr)$. $\endgroup$
    – Bernard
    Oct 26, 2017 at 23:10

3 Answers 3

3
$\begingroup$

Sorry to burst your bubble, but you're wrong about $$\frac{1}{n\ln(n)}\lt \frac{1}{n^{1.1}}$$ Notice that for positive integer $n$, this statement is equivalent to $$\ln(n)\gt n^{0.1}$$ First of all, this is false for $n=1,2,3$, but that doesn't matter, since we're looking at asymptotic domination here. So we should instead consider whether or not $$\ln(n)\gt n^{0.1}$$ is true for large $n$. If this is true for large $n$, then the value of the limit $$\lim_{n\to\infty} \frac{\ln(n)}{n^{0.1}}$$ should be $\infty$. However, by L'Hopital, $$\lim_{n\to\infty} \frac{\ln(n)}{n^{0.1}}=\lim_{n\to\infty} \frac{\frac{1}{n}}{0.1n^{-0.9}}$$ $$\lim_{n\to\infty} \frac{\ln(n)}{n^{0.1}}=\lim_{n\to\infty} \frac{10}{n^{0.1}}$$ $$\color{red}{\lim_{n\to\infty} \frac{\ln(n)}{n^{0.1}}=0}$$ Which disproves your claim that $$\frac{1}{n\ln(n)}\lt \frac{1}{n^{1.1}}$$

$\endgroup$
2
  • $\begingroup$ Thank you!! I was sure that the inequality was wrong but I didn't know why. Thank you for clarifying $\endgroup$
    – J. Door
    Oct 26, 2017 at 23:26
  • $\begingroup$ @J.Door No problem! $\endgroup$ Oct 26, 2017 at 23:28
1
$\begingroup$

Note the inasmuch as $\log(n)\le n-1<n$, and $\log(n^{0.1})=0.1\log(n)$, we see that

$$\sum_{n=2}^N \frac1{n\log(n)}\ge 0.1\sum_{n=2}^N \frac{1}{n^{1.1}}\tag 1$$

One cannot conclude from $(1)$ that the partial sums on the left-hand side converges (they do not) as a consequence of convergence of the right-hand side of $(1)$.

$\endgroup$
0
$\begingroup$

Perhaps it is more clear to consider $n=e^{x_n}$ so that we get

$$\frac1{\ln(n)}=\frac1{x_n}$$

$$\frac1{n^{0.1}}=\frac1{e^{0.1x_n}}$$

$$\frac1{x_n}\quad\text{vs.}\quad\frac1{e^{0.1x_n}}$$

Which one converges to zero faster? Hopefully you realize exponential growth dominates linear growth.

$\endgroup$
1
  • $\begingroup$ @MarkViola Whoops, typo. $\endgroup$ Oct 27, 2017 at 13:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .