# Does the following series (inspired by harmonic things) converge?

$$\left[e - \left(1+ \left(\frac{1}{\frac{1}{2} + \frac{1}{3} + \frac{1}{4}} \right) \right) ^ {\left(\frac{1}{2} + \frac{1}{3} + \frac{1}{4} \right)} \right] +$$

$$\left[e - \left(1+ \left(\frac{1}{\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}} \right) \right) ^ {\left(\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} \right)} \right] +$$

$$\left[e - \left(1+ \left(\frac{1}{\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} +\frac{1}{6}} \right) \right) ^ {\left(\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6}\right)} \right] + ...$$

• Could you add to your post why you think it might converge, or what led you to this series? Feb 19, 2020 at 2:58
• I was thinking about $\frac{1}{2}, \frac{1}{2+ \frac{1}{2}}, \frac{1}{2+ \frac{1}{2} +\frac{1}{3}}, \frac{1}{2+ \frac{1}{2} +\frac{1}{3} +\frac{1}{4}}$ converging to zero more slowly than $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, ...$ and was wondering whether I could use the more slowly converging sequence to create a simple problem. The rest is explained by Milo Brandt in his answer: "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." Feb 19, 2020 at 3:48

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.
• "that denominator is known to be asymptotic to $log(𝑛)$" suggests that if $t_k = \frac{1}{k}$ and we want a simple example of a function f such that ... $$\lim_{n\to \infty} \left( \int_{ \sum_{k=1}^n t_k}^{\sum_{k=1}^{n+1} t_k} f(x) dx \right) = 1$$ then it suffices to have $f(log (y)) =y$ Feb 19, 2020 at 4:43