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2 votes
1 answer
189 views

Bound on a two-dimensional recursive series

For $n,k\in\mathbb{N}$, let $f(n,k)$ be defined as follows. If $n \geq k$ and $n > 2$, then $$ f(n,k) = \frac{k(n-k)}{n(n-1)}f(n-2,k-1) + \frac{k(k-1)}{n(n-1)}f(n-2,k-2) + \frac{n-k}{n}f(n-1,k) + \...
macat's user avatar
  • 155
2 votes
1 answer
349 views

Lower bound on sum of independent heavy-tailed random variables

I have a sum of $n$ i.i.d random variables $X_i$ such that $E[X_i] = 0$,$\mathrm{E}[|X_i|^{1 + \delta}]$ exists for some $0 < \delta < 1$ but $\mathrm{E}[|X_i|^{1 + \delta+ \epsilon}]$ does not ...
Kaiyue Wen's user avatar
2 votes
1 answer
383 views

Lower bound and limit of a sum with binomial coefficients

Let $$A_k = \sum_{i=1}^k i {3k-2i-1 \choose i-1} {2i-2 \choose k-i}$$ $$B_k = \sum_{i=1}^k i {3k-2i-2 \choose i-1} {2i-1 \choose k-i}$$ $$C_k = \sum_{i=1}^k (3k-2i-2) {3k-2i-3 \choose i-1} {2i\...
macat's user avatar
  • 155
5 votes
4 answers
917 views

Limit of a sum with binomial coefficients

Let $$A_k = \frac{\sum_{i=1}^ki{2k-i-1 \choose i-1}{i-1 \choose k-i}}{k{2k-1\choose k}}$$ $$B_k = \frac{\sum_{i=1}^ki{2k-i-2 \choose i-1}{i \choose k-i}}{k{2k-1\choose k}}$$ $$C_k = \frac{\sum_{i=1}^k(...
macat's user avatar
  • 155
11 votes
3 answers
1k views

What is the limit of $a (n + 1) / a (n)$?

Let $a(n) = f(n,n)$ where $f(m,n) = 1$ if $m < 2 $ or $ n < 2$ and $f(m,n) = f(m-1,n-1) + f(m-1,n-2) + 2 f(m-2,n-1)$ otherwise. What is the limit of $a(n + 1) / a (n)$? $(2.71...)$
José María Grau Ribas's user avatar
0 votes
0 answers
64 views

if $\max_{z \in K} |\zeta(z+it)-f(z)|<\epsilon.$ then is this $\lim_{t\to \infty} \inf \frac {|\zeta'(z+it)|}{|f'(z)|} $ a finit limit?

Universality theorem of Riemann zeta function states that :Let $K$ be a compact subset with connected complement lying in the strip $\{1/2 < \operatorname{Re}(z)<1\}$, and let $f : K \rightarrow ...
zeraoulia rafik's user avatar
3 votes
2 answers
283 views

Lower bound for Euler's function

Euler function is defined, for $|x|\le 1$, as follows: $$\phi(x)=\prod_{i=1}^\infty(1-x^i)$$ Upper bounds for $\phi$ can be simply derived from ending the product early, e.g. $$\phi(x)<\prod_{i=1}^...
R B's user avatar
  • 618