# If $\lim \limits_{n \to \infty} nb_n = 0$, construct a convergent series such that $\lim \limits_{n \to \infty} \frac{b_n}{a_n} =0$.

The question asks to prove that if $b_n = o(\frac 1n)$ as $n \to \infty$, one can always construct a convergent series $\sum_{n=1}^{\infty} a_n$ such that $b_n = o(a_n)$ as $n \to \infty$.

What I have tried so far:

If I choose an $N \in \Bbb N$ such that $|nb_n|<1$ for all $n>N$, then $b_n < \frac 1n$. By the completeness axiom I can choose an $a_n$ such that $b_n < a_n < \frac 1n$. With some $a_n$ within these bounds for each $n$, $$0 < \left| \frac {b_n}{a_n} \right| < \left|\frac{b_n}{\left (\frac 1n\right)}\right| = |\ nb_n|$$

for all $n> N$ so $\lim \limits_{n \to \infty} \frac{a_n}{b_n} = 0.$

At this point I'm stuck as I don't know if what I've done is valid or how to formulate $a_n$ more precisely if it is. Also, showing that the limit above converges to 0 seems like a circular argument.

However, since $\sum_{n=1}^{\infty} \frac{1}{n^p}$ does converge for all $p > 1$, I should be able construct $a_n$ so that the sum converges using the comparison theorem and ensuring each $a_n$ is non negative.

Edit: looking at the question again, it does not specify that $b_n = o( \frac 1n)$ for the base $n \to \infty$. But since it gives no other base I assume this one was implied.

• Your inequality manipulation is suspicious. If $a_n<\dfrac1n$ then $\dfrac1{a_n}> n$. Commented Mar 31, 2014 at 14:56
• Surely just put $a_n:= b_n/n^2$ which converges by comparison with $\sum 1/n^2$? Commented Mar 31, 2014 at 15:03
• $b_n/a_n=n^2$... Commented Mar 31, 2014 at 15:04
• @TedShifrin whoops! I'm not sure if I can remedy that... Commented Mar 31, 2014 at 15:04
• The biggest problem for me was that the sum of $b_n$ does not necessarily converge but somehow $a_n$ has to converge and decay more slowly than $b_n$ Commented Mar 31, 2014 at 15:06

Let $$B_n=\sup_{k\ge n}|k\,b_k|.$$ Then $B_n$ is non increasing, $|n\,b_n|\le B_n$ and $\lim_{n\to\infty}B_n=0$. Let $a_n=(-1)^nB_n$. By Leibniz's criterion $\sum a_n$ is convergent, and $$\Bigl|\frac{b_n}{a_n}\Bigr|=\frac{|b_n|}{B_n}\le\frac{1}{n}.$$