I have just read about two eggs problem. I know that with decreasing amount of jumps we can reach worst case scenario of first jump $a = \sqrt{2n}$, $n$ is the number of floors, how about the lower bound of the problem?
Let me make it more precise about the issue that I am having. There are two cases: 1. regular jumps
with regular jumps, let's say it is a jumps per probe
$T(k,n) \leq n/a + T(k-1,a)$ for worst case, where $k$ is number of eggs and $n$ is number of floors
$T(2,n) \leq 2\sqrt{n}$
$T(3,n) \leq 3n^{1/3}$
By induction, $T(k,n) \leq kn^{1/k}$
How about its lower bound? 1. gradually smaller jumps
$a +(a-1)+(a-2)+\dots+1 = n$
so $a =\sqrt{2n}$.
So what is the upper and lower bound for this case?
This is where I read about it: https://www.cs.umd.edu/~gordon/ysp/egg.pdf
All you have to do is to understand the idea in the Wikipedia article about this very problem, which is that it is not good to use the parametrization in your question. Namely, instead of looking for the minimum number of tries needed given $n$ eggs and $k$ floors, you look for the maximum number of floors testable given $n$ eggs and $t$ tries, which follows the trivial recurrence relation. Prove that this is monotonic in $t$, so that you can use any root-finding algorithm to find the minimum $t$ given $n,k$. The article mentions binary search because that's the simplest and you can't do much better than a log factor in practice anyway.
Specifically, following the article by letting $f(t,n)$ be the maximum number of floors testable with $t$ tries and $n$ eggs, you get $f(t,2) = \binom{t}{0} + \binom{t}{1} + \binom{t}{2}$, and it is trivial to invert this function to find the minimum $t > 0$ such that $f(t,2) \ge k$, where $k$ is the number of floors, since it is just a quadratic. Deal with small cases separately where the quadratic is nonzero everywhere. There is no need for upper/lower bounds since the answer is exact. For higher $n$, if you just want an approximation, simply approximate $f(t,n)$ before inverting. For example, one bound is $\binom{t}{3} \le f(t,3) \le \binom{t+3}{3}$, which allows you to find the optimal $t$ to within $3$ of the correct answer if you can invert $x \mapsto \binom{x}{3}$. I leave it up to you to find other more convenient bounds, such as $c(t-a)^3 \le f(t,n) \le c(t+b)^3$ for some constants $a,b,c$, so that inverting is easier.