Comparison Test prove $b_n \le a_n$

Question

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?

Answer 1

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 $n\ge1$.

Answer 2

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.

Answer 3

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 $a_n$.