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Let $x$ be real and define the entire function $f(x)$ as

$$ f(x) = 1 + \sum_{n=1}^{\infty} (\frac{x}{\ln^2(n+1)})^n $$

Now we have that

$$ \lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} (1 + \sum_{n=1}^{\infty} (\frac{x}{\ln^2(n+1)})^n) = 0$$

where the limit is approached from below.

How to prove that ?

edit : some background, source and inspiration for those who are curious

A certain integral idea might perhaps help, see the similar question :

Solve for $m(t)$ in the integral transform $\int_0^1(1-t^n) m(t) dt=\frac{(n+1)^n}{n^{n-1}} $ for $n>0$.

On the other hand this function $f(x)$ is not strict positive.

It more resembles the question here

$0 \leq C + \sum_{n=1}^{+\infty} \frac{x^n}{(n+1)^n} $ ,where $C = 1.036055393...$ is the best possible constant.

But it seems a bit harder.

edit 2

I found some arguments :

Consider the function $\exp(\exp(x)-1)$.

They have the taylor coefficients :

$$\frac{B_n}{n!}$$

and those are bounded by

$$ (\frac{c_0}{\ln(n+1)})^n < \frac{B_n}{n!} < (\frac{c_1}{\ln(n+1)})^n$$

for some constants $c_0,c_1$.

Therefore looking similar to the function $f(x)$.

In fact,

For $x>>1$ we have the estimate $f(x) = x^{t(x)}$ where $t(x)$ is the functional inverse of $\ln(x+1)^{2x}$. This implies that $f(x)$ for large positive $x$ grows about double exponentially. Similar to Bell and $\exp(\exp(x)-1)$ thus.

We know that double exponentials have a limit at minus infinity.

So that makes it a bit less surprising.

Then again this $f(x)$ grows a bit slower... And imo that makes it interesting.

Notice $\frac{x}{1-x} = 1 + \sum_n x^n $ also has a limit as $x$ goes to $- \infty$.

This function seems somewhat " sandwiched " between the double exp and this rational function, although that is ofcourse NOT a formal argument. (do not confuse with the sandwich theorem).

mick
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