0
$\begingroup$

On my calculus II exam, my professor wanted us to determine whether the below series was convergent or divergent. $$\sum _{n=1}^{\infty }\:\frac{1}{\sqrt[4]{n^3+1}}$$ I realized that it was most likely divergent, so I used the comparison test, trying to find a smaller series $b_n$ that was also divergent. $$a_n = \frac{1}{\sqrt[4]{n^3+1}}$$ $$b_n = \frac{1}{n^{\frac{4}{5}}}$$ He took off a bunch of points because he said $b_n$ was not $\le a_n$.

I would like to prove to him that $$b_n\le a_n$$ What would you recommend I do?

$\endgroup$
  • $\begingroup$ Welcome to Maths SX! Is the exponent in $b_n$ really $4/5$? $\endgroup$ – Bernard Apr 16 at 20:56
  • $\begingroup$ Yes it was 4/5 @ Bernard. @lulu I was less sure about 3/4 for some reason $\endgroup$ – Zeke Legge Apr 16 at 20:58
  • $\begingroup$ Well, $\frac 34$ is the way to go. Sure $a_n<b_n$ with that choice, but the limit of their ratio is $1$. Or you could work with $\frac 1{2n^{3/4}}$. $\endgroup$ – lulu Apr 16 at 20:58
  • $\begingroup$ @lulu unfortunately I forgot how to perform the LCT during the exam $\endgroup$ – Zeke Legge Apr 16 at 20:59
  • $\begingroup$ Do you know what equivalent functions are, in asymptotic calculus? $\endgroup$ – Bernard Apr 16 at 21:00
0
$\begingroup$

I will speculate that your professor is penalizing you for being mathematically imprecise. (If I were your professor that's what I'd be inclined to do. Be glad I'm not your professor.) That is, in comparing $\sum a_n$ to $\sum b_n$, you are tacitly saying that $\sum_{n=1}^\infty a_n\ge\sum_{n=1}^\infty b_n$ because $a_n\ge b_n$ for all $n$. But that is not true. In particular $a_1=1/\sqrt[4]2$ is less than $b_1=1$.

There is, of course, an easy fix: Just say that $\sum a_n$ diverges because $\sum b_n$ diverges and $a_n\ge b_n$ for all $n$ greater than or equal to $2$. I'm not sure that that would completely mollify your professor, but at least it would be an indication that you've given some thought to why the inequality is true.

Proving the inequality is true for $n\ge2$ is actually a little delicate, since $1/2^{4/5}\approx0.574$ is awfully close to $1/\sqrt[4]{2^3+1}=1/\sqrt3\approx0.577$. One proof begins something like this:

$${1\over n^{4/5}}\le{1\over\sqrt[4]{n^3+1}}\iff(n^3+1)^5\le n^{16}\iff\cdots$$

At this point I would be tempted to abandon the attempt to prove the inequality all the way down to $n=2$ and instead say

$$n\ge32\implies n^{16}\ge32n^{15}=(n^3+n^3)^5\ge(n^3+1)^5\implies{1\over\sqrt[4]{n^3+1}}\ge{1\over n^{4/5}}$$

and conclude that $\sum a_n$ diverges because $\sum b_n$ diverges and $a_n\ge b_n$ for all $n\ge32$.

But what I really suggest is instead to compare $a_n=1/\sqrt[4]{n^3+1}$ to $b_n={1\over2n}$. The proof of the relevant inequality is straightforward:

$${1\over2n}\le{1\over\sqrt[4]{n^3+1}}\iff n^3+1\le16n^4\iff1\le(16n-1)n^3$$

and the rightmost inequality is obviously true for $n\ge1$.

$\endgroup$
  • $\begingroup$ For the comparison rest it suffices that $\{n\in \Bbb N: a_n>b_n\}$ is finite, i.e. bounded. However if the proposer wrote on the test that $b_n\le a_n$ for all $n\in \Bbb N$ then a deduction of marks is justifiable....To the proposer: For all $n \in \Bbb N$ we have $a_n\ge (n^3+n^3)^{-1/4}=2^{-1/4}n^{-3/4}\ge 2^{-1/4}n^{-1}.$ $\endgroup$ – DanielWainfleet Apr 17 at 1:01
0
$\begingroup$

So here a very fast way to go, with equivalents. Remember two functions are asymptotically equivalent if their ratio tends to $1$ at $\infty$.

For instance, a polynomial is asymptotically equivalent to its leading term. The main rule to know is that equivalence is compatible with multiplication and division, but not with addition nor subtraction. Also, a function which has a finite limit $\ell$ at $\infty$ is asymptotically equivalent to $\ell$, and conversely.

The theorem we use is that two series with positive asymptotically equivalent terms both converge or both diverge.

In the case at hand, we have $\;n^3+1\sim_\infty n^3$, hence $\;\dfrac{1}{\sqrt[4]{n^3+1}}\sim_\infty \dfrac 1{n^{3/4}}$, and the latter diverges.

$\endgroup$
  • $\begingroup$ Very interesting. It all hinges on dismissing the additive 1 because it is an infinite series, which is intuitive enough. $\endgroup$ – Zeke Legge Apr 16 at 21:15
  • $\begingroup$ Even a term in, say, $n^2$ could be dismissed. $\endgroup$ – Bernard Apr 16 at 21:18
0
$\begingroup$

You’re right. Observe:

$\lim_{n \rightarrow \infty} \frac{b_n}{a_n} = \ lim_{n \rightarrow \infty} n^{-1/20} = 0$

Which implies that$b_n$ becomes asymptotically slower than $a_n$.

$\endgroup$
  • $\begingroup$ I am not debating you, but the graph of both functions suggests to me that b_n is smaller over the interval [1,infinity) photos.app.goo.gl/1EvrwagieYGUBNoL9 Does your solution take the interval into account? @gdepaul $\endgroup$ – Zeke Legge Apr 16 at 21:06
  • 1
    $\begingroup$ The ratio $b_n/a_n$ is asymptotically $n^{-1/20}$, not $n^{1/20}$. $\endgroup$ – Barry Cipra Apr 16 at 21:51
  • $\begingroup$ Lol, my bad . I made a calc error on my end. @ZekeLegge you are right, but i think your professor actually wanted you to perform the calculation on n^{-3/4} instead of n^{-4/5} $\endgroup$ – gdepaul Apr 16 at 22:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.