4
votes
Asymptotic of $\sum_{k=0}^\infty\frac{\Gamma (k+n+1) \Gamma (3 k+n+1)}{\Gamma (k+2) \Gamma (3 k+2 n+2)}$ as $n\to \infty$
First of all we note that
$$g(n)=\sum_{k=0}^{n/2}\frac{\Gamma (a+k+n+1) \Gamma (b+3 k+n+1)}{\Gamma (c+k+2) \Gamma (d+3 k+2 n+2)}\ll\sum_{k=n/2}^\infty\frac{\Gamma (a+k+n+1) \Gamma (b+3 k+n+1)}{\Gamma (...
- 13.9k
1
vote
Does this recursive journey through Pascal's triangle always reach $1$?
At least heuristically, there should be an $a_1$ for which the $a_n$ tend to infinity. First of all, we can take $a_1$ large enough to consider only asymptotics. Now, note that $p(k)$ 'looks like' ${\...
- 51.4k
1
vote
Is there an alternative proof for $\lim a_n^{\frac{1}{q}}=(\lim a_n)^{\frac1q}$?
The function $\exp: \mathbb{R} \to \mathbb{R}^{>0}$ and its inverse $\log : \mathbb{R}^{>0} \to \mathbb{R}$ are continuous. Clearly $x \to x/q$ is also continuous on $\mathbb{R}^{>0}$. Thus $...
- 24.9k
1
vote
Accepted
Prove that $\lim_{z \rightarrow 1-i} |\bar{z}^{2}-2| = 2\sqrt{2}$
If your intention is to solve it through $\epsilon - \delta$ definition, then observe that:
$$|(z^2-2)-((1-i)^2-2)| =|z^2-(1-i)^2|.$$
Now,
$$|z^2-(1-i)^2| = |z-(1-i)||z+(1-i)|.$$
That is
$$|z^2-(1-i)^...
- 1,528
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