Example Sequence $\{ b_n \}$ with $\lim_{n \rightarrow \infty} nb_n = 0$ but $\sum_{n=1}^{\infty} b_n$ diverges.

Question

As the title states, I'm looking for an example of a strictly decreasing sequence of positive numbers with the properties that $$ \lim_{n \rightarrow \infty} nb_n = 0$$ but $$ \sum_{n=1}^{\infty} b_n $$ diverges.

My efforts have been unsuccessful so far. I know that nothing of the form

$$ b_n = \frac{1}{n^p} $$

works, as $\lim_{n \rightarrow \infty} nb_n = 0$ if $p>1$, also implying that the series will converge via p-test. I've also tried more creative sequences like $$ b_n = \frac{\sin(\frac{1}{n})}{n} $$ but still no luck.

More than a specific example, is there a certain strategy I should employ to find such an example? I was thinking the sequence must go to zero must faster than $n$ goes to infinity, but not fast enough for the series to converge.

Answer 1

Hint :

Look at Bertrand series.

Bertrand series are the series of the form : $$\frac{1}{n^{\alpha}(\log n)^{\beta}}$$ and you know that this serie converges only if : $\alpha > 1$ or ($\alpha = 1$ and $\beta > 1$).

Answer 2

With $$s_n:=\sum_{k=1}^n b_k,$$ try to achieve that $s_n\to \infty$, but sloooowly.

If $s_n= n$, then $b_n=1$ which is too large.

If $s_n=\sqrt n$, then $b_n\sim \frac1{\sqrt n}$, which makes $nb_n\sim\sqrt n$, still too large.

If $s_n=\ln n$, then $b_n\sim \frac 1n$, still too large: $nb_n\sim 1$.

In general, if $s_n=f(n)$ then $b_n\sim f'(n)$.

Now what if $s_n=\ln\ln n$? Then $b_n\sim \frac1{n\ln n}$ and $nb_n\sim \frac1{\ln n}\to 0$!

Answer 3

Let $p_n$ denote the $n$-th prime number. Then $$\sum_{n\in\Bbb N}\frac{1}{p_n}=\infty\tag{1}$$

But $$\lim_{n\to\infty}\frac{n}{p_n}=0\tag{2}$$

For $(1)$, see Divergence of the sum of the reciprocals of the primes.

For $(2)$, this is the Prime Number Theorem: $p_n\sim n\log n$. See for instance this discussion. This is also another way to prove $(1)$.

Answer 4

Using the idea in this answer: Is there a slowest rate of divergence of a series?

We set $D_n = 1/n$, so $\sum_{n = 1}^{\infty}D_n$ converges. Write $H_n = \sum_{k = 1}^{n}D_n$, the $n$th harmonic number. Finally, let $d_n = D_n/H_{n-1} = \frac{1}{nH_{n-1}}$. We can easily show that $d_n$ is positive and decreasing and $nd_n \rightarrow 0$, and that answer shows that $\sum_{n = 2}^{\infty}d_n$ diverges.

Asymptotically this answer is no different from $b_n = \frac{1}{n\ln(n)}$ given in other responses, but I thought this would be a nice fact to share.