# A conjecture relating an integral and a sum, the floor function and squares

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.

• Your subject line said that you were asking for help, and your last paragraph asked people not to get angry. I think neither of these adds to the problem—a question might or might not be well received, but I would say this community almost never gets angry at someone asking a question in good faith. I have edited them out, but, if you are sure that they add to the problem, please feel free to edit them back in. (Introducing $x$, and most of your last paragraph, seem redundant; it seems to say no more than the equality already asserted. If you agree, then you might edit that, too.) Jan 26 at 21:08
• Thank you I appreciate your edits. The last sentence was to mostly state that C tends to 1, as x goes to infinity, not necessarily to state again that there would be a remainder less than 1. But, I am open to all suggestions and edits frankly. I accidentally posted this on math.se and thats the wrong place for conjecture so I'm a little shy.
– user475930
Jan 26 at 21:13
• for $x=1$ I obtain 1.0387 for the integral and 0.971499 for the sum, so their integer part is not equal --- you say that this is an issue with numerical precision? Jan 26 at 21:19
• My apologies, I forgot to include the stipulation that x is greater than 1. I transcribed this from another document.
– user475930
Jan 26 at 21:40

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.

• By the way, Wolfram Alpha says that already for $t = 1.01$ we have $\sum_{n = 2}^{\infty} \frac{1}{(n t - 1)^{n + 1}} > 1$, that is already for $x = 10^4$ the floor of the integral is equal to $x + 1$. Jan 26 at 21:41
• @TheHoyt, whether or not begging is appropriate, you made 5 requests in about 20 minutes. @‍Random may or may not help, but at least it is appropriate to give them the courtesy of a bit of time to consider and respond, if they choose to do so. Jan 26 at 22:26
• Thank you for not spoiling the fun :D That was supremely fun.
– user475930
Feb 3 at 3:00
• The computed limit for the sum is x + 0.9885435... and is terribly hard to compute due to needing the Stieltje constants. Going to make a new question on here with my 'proof' of the sum.
– user475930
Feb 7 at 15:53
• That series1/(n-1)^n+1 is in the estimation in the sum as well, but there are an infinite amount of correction terms that bring it down below one, mainly because Stieltjes(1) is negative.
– user475930
Feb 7 at 19:06