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# Finding the limit $\lim_{x\to 0^+}e^{1/x}\sum_{n=\lfloor 1/x\rfloor}^\infty\frac{x^n}{n}$

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Let $\lfloor x \rfloor$ be the maximum integer $n\le x$. Find the limit $$\lim_{x\to 0^+}e^{1/x}\sum_{n=\lfloor 1/x\rfloor}^\infty\frac{x^n}{n}$$

I do not have an idea how to approach this problem except for a few observations. I don't have much progress from here:

• The limit is of the form $\infty\cdot 0$.
• Integrating the power series for $\frac{1}{1-x}$ term by term, one has $\sum_{n=1}^\infty\frac{x^n}{n} = -\log(1-x)$ for small positive $x$ near $0$.
• The expression can be rewritten as $$\lim_{x\to 0^+}e^{1/x}\left(-\log(1-x)- \sum^{\lfloor 1/x\rfloor}_{n=1}\frac{x^n}{n} \right)$$
• Or as $$\lim_{x\to 0^+}\sum_{n=0}^\infty\frac{1}{n!x^n}\sum_{n=\lfloor 1/x\rfloor}^\infty\frac{x^n}{n}$$
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First, let's make things simpler and do a change of variable, replacing $x$ with $\frac{1}{x}$:

$$\lim_{x\to \infty}e^x\sum_{n=\lfloor x\rfloor}^\infty\frac{1}{nx^n}$$

Massage into an explicit L'Hôpital indeterminate form:

$$\lim_{x\to \infty}\frac{\sum_{n=\lfloor x \rfloor}^\infty\frac{1}{nx^n}}{e^{-x}}$$

Now we have some work to do before we can apply L'Hôpital's rule, as the numerator is not differentiable (it's not even continuous!). The trick here is to get a pair of differentiable functions that bound the numerator and result in the fraction converging to the same value, and use those instead.

This is where I reach into a bag of analysis tricks and pull out the Lerch transcendent $\Phi$.

\begin{align} \Phi\left(\frac{1}{x}, 1, x\right) &= \sum_{n=0}^\infty\frac {1} {x^n(n+x)} \\\\ &=\sum_{n=x}^\infty\frac{1}{nx^{n - x}}\qquad\text{for integer x} \end{align}

This is very close to what we want! We can bound our numerator below with $\frac{1}{x^x}\Phi\left(\frac{1}{x}, 1, x\right)$:

$$\lim_{x\to \infty}\frac{\frac{1}{x^x}\Phi\left(\frac{1}{x}, 1, x\right)}{e^{-x}}$$

And we can bound it above with $\frac{1}{x^x}\Phi\left(\frac{1}{x}, 1, x\right) + \frac{1}{(x - 1)x^{x - 1}}$. In the limit, the second term here isn't going to matter—you can repeat the entire analysis below with this as the numerator and see that both substitutions will converge to the same value. To keep things simple, I'll only do the analysis for the first substitution.

Again, we have done this because $\Phi$ is differentiable—as the above Wikipedia page indicates, it has an integral form. We can now check the L'Hôpital rule criteria:

1. Indeterminacy of form: Both numerator and denominator go to 0 in the limit.
2. Differentiability of functions: the derivative of the numerator is (and I confess to using Wolfram Alpha at this point):
$$-x^{-x}\left(\Phi\left(\frac{1}{x}, 2, x\right) + \Phi\left(\frac{1}{x}, 1, x\right) \log(x) + \frac{1}{x - 1}\right)$$

And of course the derivative of the denominator is $-e^{-x}$.

1. Non-zero derivative of denominator: $-e^{-x}$ is never 0.

2. Existence of limit of the quotient of the derivatives: Let's go:

\begin{align} &\lim_{x\to\infty} \frac{-x^{-x}\left(\Phi\left(\frac{1}{x}, 2, x\right) + \Phi\left(\frac{1}{x}, 1, x\right) \log(x) + \frac{1}{x - 1}\right)}{-e^{-x}} \\\\ &\qquad=\lim_{x\to\infty} \left(\frac{e}{x}\right)^x\left(\Phi\left(\frac{1}{x}, 2, x\right) + \Phi\left(\frac{1}{x}, 1, x\right)\log(x) + \frac{1}{x - 1}\right) \\\\ \end{align}

The only part of this that doesn't obviously converge to 0 as $x$ grows is $\Phi\left(\frac{1}{x}, 1, x\right) \log(x)$, which expands out to $$\sum_{n=0}^\infty \frac{\log(x)}{x^n(n + x)}$$ This converges to 0 because $\log(x)$ grows slower than $x^n$ for any positive $n$.

So our substitute function converges to 0, as does the original expression.

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