I've found through evidence and have conjectured on a math publication that:
$$\Big\lfloor\int_1^\infty (k^{1/(k^{1+1/\sqrt{x}})} – 1)dk\Big\rfloor = \Big\lfloor\sum_{k=1}^{\infty}k^{1/(k^{1+1/\sqrt{x}})} -1\Big\rfloor = x $$
where $ x \in \mathbb{N}, x>1$.
It is very hard to compute these values. Repeated Shanks transformations and Richardson's Extrapolation will be required to compute, or using Pari GP techniques. Before you post a counter example below 10^7 for the sum, please check your precision.
Proving this has proved extremely difficult.
My question is, does anyone have any suggestions of how to prove this? The only information I have is that this is true from all tests for $x$ less than 10^7 and we're still running tests for the sums.
They aren't equal without the floor function, and each equal $x + C$, where $C$ is a constant less than 1, and $C$ is different for the integral and sum. As $x$ tends to infinity, $C$ tends to 1.
Best Answer
First of all, $$k^{1/k^t}=e^{(\log k)/k^t} = \sum_{n = 0}^{\infty} \frac{\left( \log k \right)^n}{n! k^{n t}}$$ and therefore $$\intop_{1}^{\infty} \left( k^{1/k^t} - 1 \right) d k = \sum_{n = 1}^{\infty} \frac{1}{n!} \intop_{1}^{\infty} \left( \log k \right)^n k^{- n t} d k =$$ $$\sum_{n = 1}^{\infty} \frac{1}{n!} \intop_{0}^{\infty} y^n e^{(1 - n t) y} d y = \sum_{n = 1}^{\infty} \frac{1}{(n t - 1)^{n + 1}}.$$ If $t = 1 + 1/\sqrt{x}$ then the main contribution to this sum is the term $n = 1$ which gives a contribution of $x$, and so the sum is equal to $$x + \sum_{n = 2}^{\infty} \frac{1}{(n t - 1)^{n + 1}}.$$ In order for the floor of this value to be equal to $x$, then we must have $\sum_{n = 2}^{\infty} \frac{1}{(n t - 1)^{n + 1}} < 1$ for each $t > 1$, or equivalently $\sum_{n = 2}^{\infty} \frac{1}{(n - 1)^{n + 1}} \leq 1$. However, this is false: indeed, the second term alone is equal to $1$, so for large enough $x$ the floor of the integral is at least $x + 1$. What is true is that the integral is always strictly less than $x + 2$, since $$\sum_{n = 3}^{\infty} \frac{1}{(n - 1)^{n + 1}} < \sum_{n = 3}^{\infty} \frac{1}{2^{n + 1}} = \frac{1}{8}.$$
Presumably, the sum can be treated in a similar manner, where the main term (which should be about $x$) would be $\sum_{k = 1}^{\infty} \frac{\log k}{k^{1 + 1/\sqrt{x}}}$, and the other terms contributing at most a constant.