Does the following series (inspired by harmonic things) converge? $$\left[e - \left(1+ \left(\frac{1}{\frac{1}{2} + \frac{1}{3} + \frac{1}{4}} \right) \right) ^ {\left(\frac{1}{2} + \frac{1}{3} + \frac{1}{4} \right)} \right] + $$
$$\left[e - \left(1+ \left(\frac{1}{\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}} \right) \right) ^ {\left(\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} \right)} \right] +$$
$$\left[e - \left(1+ \left(\frac{1}{\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} +\frac{1}{6}} \right) \right) ^ {\left(\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6}\right)} \right] + ...$$

This series diverges, but it's somewhat instructive in why - this is an instance where we can just "peel back" all the layers of the expression by applying linear approximations and eventually reduce it to something trivial.
In particular, I assume that you know
$$\lim_{x\rightarrow 0}(1+x)^{1/x} = e$$
and thought to look at the rate of convergence of this, essentially, by looking at what happens if we approach $x=0$ fairly slowly and sum up the error terms. The trouble is that the behavior of the function $f(x)=(1+x)^{1/x}$ actually isn't all that interesting: it's some smooth function with non-zero derivative at $0$. Although it's not that important, the derivative turns out to be $-e/2$ at $0$ which means that if $\alpha < -e/2 < \beta$ then, for all small enough $x$ we have that $f(x)-e$ is between $\alpha x$ and $\beta x$ - so is essentially linear.
In particular, this suffices to say that if $x_n$ is a sequence with $0$ as first limit, then $\sum f(x_n)$ converges if and only if $\sum x_n$ does. Okay, so now we've unravelled the outermost detail just by noticing that "differentiable" is basically as good as "linear" here, which is not interesting for the matter of convergence.
So, let's define
$$x_n=\frac{1}{\frac{1}2 + \frac{1}3 + \ldots + \frac{1}n}$$
This, again, is easy enough to handle: that denominator is known to be asymptotic to $\log(n)$ - meaning, again, that it is always within ratio of $\log(n)$ for all large enough $n$. This can be derived via comparison with integrals of $\frac{1}x$. So, again, $\sum x_n$ is going to converge if and only if $\sum \frac{1}{\log(n)}$ converges.
At this point we can stop: $\frac{1}{\log(n)}$ is bigger than $\frac{1}n$, and we know that the harmonic series diverges (indeed we just used an asymptotic for it!), so clearly $\sum \frac{1}{\log(n)}$ diverges. Since the partial sums of this turn out to always be within a constant factor of those of the original, the original sum diverges.
