Prove divergence and conclude that there is no universal "smallest" comparison series to test divergence

This is a practice problem for a midterm in a real analysis undergrad class, but I tagged it as homework anyway.

Suppose that $a_k>0$ and that the series $\sum_{k=1}^\infty a_k$ diverges. Let $S_n$ = $\sum_{k=1}^n a_k$ and define $b_1 = a_1$ and $b_n = \sqrt{S_n}-\sqrt{S_{n-1}}$ for $n\ge2$. Prove that $\sum_{k=1}^\infty b_k$ diverges, and that $\lim_{x\to\infty}\frac{b_n}{a_n}=0$. Conclude that there is no universal "smallest" comparison series to test divergence.

So $S_n = S_{n-1}+a_n$ and as n increases that $a_n$ will diverge. The square roots are messing me up, I don't know how to start with the limit and I understand the last statement, but I don't see how I can conclude it.

Solution 1:

These are some hints.

To prove that $\sum\limits_nb_n$ diverges, use the fact that $b_n=c_n-c_{n-1}$ for some sequence $(c_n)_n$ and express the partial sums $\sum\limits_{n\leqslant N}b_n$ in terms of $(c_n)_n$.

To prove that $b_n/a_n\to0$, rewrite $b_n$ as a ratio, multiplying and dividing $b_n$ by the conjugate quantity $\sqrt{S_n}+\sqrt{S_{n-1}}$.

The conclusion is that for every divergent series $\sum\limits_na_n$, there exists another divergent series $\sum\limits_nb_n$ such that $b_n=o(a_n)$, in this sense the latter is smaller than the former. In particular there exists no sequence $(t_n)_n$ such that $\sum\limits_na_n$ diverges if and only if $a_n=\Omega(t_n)$.

Solution 2:

To show that $\sum_{k=1}^\infty b_k$ diverges, you clearly have to look at the partial sums. As you might expect from the definition of $b_n$, it turns out that they telescope:

$$\begin{align*} &\sum_{k=1}^nb_k=a_1+\sum_{k=2}^nb_k=a_1+\sum_{k=2}^n(\sqrt{S_k}-\sqrt{S_{k-1}})=\\ &=a_1+\sum_{k=2}^n\sqrt{S_k}-\sum_{k=2}^n\sqrt{S_{k-1}}=a_1+\sum_{k=2}^n\sqrt{S_k}-\sum_{k=1}^{n-1}\sqrt{S_k}=\\ &=a_1+\sqrt{S_n}-\sqrt{S_1}=a_1+\sqrt{S_n}-\sqrt{a_1}\;. \end{align*}$$

Since $S_n\to\infty$ as $n\to\infty$, the same is true of the $\sqrt{S_n}$’s, and it follows that $\sum_{k=1}^\infty b_k$ diverges.

Now for $n\ge 2$ we have $$\begin{align*} \frac{b_n}{a_n}&=\frac{\sqrt{S_n}-\sqrt{S_{n-1}}}{a_n}\\ &=\frac{\sqrt{S_n}-\sqrt{S_{n-1}}}{a_n}\cdot\frac{\sqrt{S_n}+\sqrt{S_{n-1}}}{\sqrt{S_n}+\sqrt{S_{n-1}}}\\ &=\frac{S_n-S_{n-1}}{a_n(\sqrt{S_n}+\sqrt{S_{n-1}})}\\ &=\frac{a_n}{a_n(\sqrt{S_n}+\sqrt{S_{n-1}})}\\ &=\frac1{\sqrt{S_n}+\sqrt{S_{n-1}}}\;, \end{align*}$$

which clearly tends to $0$ as $n\to\infty$. (When you have a difference of square roots, it’s always worth thinking about this trick.)

Finally, the fact that $\frac{b_n}{a_n}$ tends to $0$ means that $\sum_{k=1}^\infty b_n$ is ‘smaller than’ $\sum_{k=1}^\infty a_n$, and yet it still diverges. That is, given a divergent series, you can always find a ‘smaller’ one, so there is no ‘smallest one’.