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?
Best Answer
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 $n\ge1$.