How to find a real sequence $x_n$ such that $n x_n \to 0$ as $n \to \infty$ and $\sum_{n=1}^{\infty} x_n$ diverges?

Question

I'm stuck on trying to find a real sequence $x_n$ such that $n x_n \to 0$ as $n \to \infty$ and $\sum_{n=1}^{\infty} x_n$ diverges. Here's what I have tried so far:

I've started with trying to find $x_n$ as a real-valued function of $n$. In order to have $n x_n \to 0$, we require $$\frac{x_n}{1/n} \to 0$$ and since the denominator tends to $0$ as $n$ increases, we require that $x_n \to 0$. We also require that $x_n$ tends to $0$ "faster" than $1/n$ does, in order for the above quotient to go to $0$.

However if you try functions such as $x_n=1/n^2$ or $x_n=n/2^n$ (for example) which tend to $0$ faster than $1/n$ does, then you see that $\sum_{n=1}^{\infty} x_n$ converges and I haven't been able to find a function for $x_n$ that satisfies the above conditions and also has a divergent infinite sum. It's certainly possible that the form for $x_n$ might not be a simple elementary function of $n$, but I don't know how you'd go about finding any "unusual constructions" that work.

Answer 1

Take $x_n = \dfrac 1 {n\ln n }.$

Answer 2

Here's one way that you might plausibly come to an answer like $x_n=\frac1{n\log n}$: We know that $\sum_{i\lt n}\frac1i\approx\log n$. If we want a divergent series, then the partial sums of our series will still need to go to infinity, but they'll need to do it at an even slower rate than $\log n$. We could try a couple of natural targets for our partial sums, say $\sqrt{\log n}$ or $\log\log n$.

But how do we get a term $x_n$ that gives us partial sums that grow this way? This is where the Euler-Maclaurin formula comes in handy: we know that we can approximate $\sum_{i=1}^n f(i)$ by $\int_1^n f(x)\ dx$, and vice versa. So we want a function $f()$ with $\int_{1}^n f(x)\ dx\approx\sqrt{\log n}$. How do we find one?

Well, we turn to another tool: the Fundamental Theorem of Calculus! Since the integral is the antiderivative, we can start with our target function $g(n)=\sqrt{\log n}$, say, and take the derivative $f(x)=\frac{dg}{dx}$ of this; by the chain rule we get $f(x)=\frac1{2x\sqrt{\log(x)}}$. The constant factor of 2 doesn't matter for convergence here, so we can multiply it away; we get that $x_n=\frac1{n\sqrt{\log n}}$ will yield a divergent series because $\sum_{i=1}^nx_i\approx C\sqrt{\log n}$ for some constant $C$. At the same time, by the way we constructed our sequence as diverging 'more slowly' to infinity than $\log n$ we have $\lim_{n\to\infty}nx_n\to 0$ (as you can see for yourself). Doing the same with $g(n)=\log\log n$ will give you the suggested $x_n=\frac1{n\log n}$ term.

You sometimes need to be a little careful going back and forth between sums and integrals like this, but for general questions of convergence and approximation it's one of the most useful tools in the toolbox.