A problem from Zorich's Analysis:
If $b_n=o(1/n)$, show that one can construct a convergent series $\sum a_n$ such that $b_n=o(a_n)$ as $n \to \infty$.
I've come up with some possible series, but can't fully prove it.
If we use $a_n=1/n$, the second condition is satisfied but not the first, so I guess I need something "smaller" than $1/n$ but "bigger" than $b_n$. My first attempts were the arithmetic or geometric mean of $b_n$ and $1/n$, but I couldn't prove their convergence nor come up with a counterexample.
Then I thought about Leibniz's criterion, and tried to set:
$$a_n=b_n-b_{n-1}+\frac{(-1)^n}{n}$$
which, according to Cauchy criterion, seem to converge. I'm not sure how to prove $b_n=o(a_n)$ though.
So, I wonder what the solution is. Also, are there any counterexamples to the previous series, that is, does the series $\sum \sqrt{|b_n|/n}$ converge?
Best Answer
You could use $ a_n = (-1)^{n+1} / n $. The series is well known to converge (with limit $1/2$).
That $b_n = o(1/n) $ as $n \to \infty $ means $|n b_n| \to 0 $. Then
$$\begin{align} \left| \frac{b_n}{a_n} \right| &= nb_n \to 0 \quad\text{as }n\to\infty \end{align}$$
so that $b_n = o(a_n) $.
I hope this helps.
Postscript: I had wondered whether $a_n$ could be found with all positive terms. I think not. If $b_n = 1/(n\log n)$, say for $n > 2$ then $b_n =o(1/n)$. However, an integral test shows that $ \sum_n b_n $ diverges, and therefore any positive series $a_n$ with $b_n =o(a_n)$ must also have divergent sum. This is not a rigorous proof but hopefully forms the basis of one.