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

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

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

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  • $\begingroup$ Could you add to your post why you think it might converge, or what led you to this series? $\endgroup$
    – coffeemath
    Feb 19, 2020 at 2:58
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    $\begingroup$ 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." $\endgroup$ Feb 19, 2020 at 3:48

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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.

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  • $\begingroup$ "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 $f(log (y)) =y$ $\endgroup$ Feb 19, 2020 at 4:43

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