Cauchy comparison test - if the series limit tends to 1 from bottom I've a following task
Check convergence of:
$$\sum_{n=2}^\infty (-1)^{\lceil 1+\sin^{2} n^{5} \rceil}\left(\frac{n^{2}+3n+10}{n^{2}+5n+17}\right)^{n^{2}(\sqrt{n+1}-\sqrt{n-1})}$$ 
My solution is:
$\lceil 1+\sin^{2} n^{5} \rceil$ is always 2, because $\forall {k\in \mathbb{Z}}$, $\sin^{2} n^{5}\neq k\Pi$, where $\Pi$ is irrational.
$$\sum_{n=2}^\infty \left(\frac{n^{2}+3n+10}{n^{2}+5n+17}\right)^{n^{2}(\sqrt{n+1}-\sqrt{n-1})}$$
Checking Cauchy test:
$$\lim_{n\rightarrow\infty} \sqrt[n]{\left(\frac{n^{2}+3n+10}{n^{2}+5n+17}\right)^{n^{2}(\sqrt{n+1}-\sqrt{n-1})}}=\lim_{n\rightarrow\infty}\left(\frac{n^{2}+3n+10}{n^{2}+5n+17}\right)^{n(\sqrt{n+1}-\sqrt{n-1})}$$
$$=\lim_{n\rightarrow\infty} \left(1+\frac{-2n-7}{n^{2}+5n+17}\right)^{n(\sqrt{n+1}-\sqrt{n-1})}=\lim_{n\rightarrow\infty}e^{\frac{-2n-7}{n^{2}+5n+17} n(\sqrt{n+1}-\sqrt{n-1})}=$$
$$=\lim_{n\rightarrow\infty}e^{\frac{(-2n-7)2n}{(n^{2}+5n+17)(\sqrt{n+1}+\sqrt{n-1})}}=-1^{-}$$
Because the series limit tends to 1 from bottom, the series converge. Is that a true assumption?

According to your calculation, there is no indication of whether the series converges or diverges.  ($|a_n^{1/n}|$ converges or diverges as $n \rightarrow \infty$ according to the limit being > or < 1, and is inconclusive if = 1.)
You made a mistake above:
$$\frac{n^2+3 n+10}{n^2+5 n+17} \approx 1 - \frac{2}{n} $$
so that
$$\left ( \frac{n^2+3 n+10}{n^2+5 n+17} \right )^{n^2 [\sqrt{n+1} - \sqrt{n-1}]} \approx \left ( 1 - \frac{2}{n} \right )^{n^{3/2}} \approx \exp{(-2\sqrt{n})} $$
as $n \rightarrow \infty$.  The above convergence test would be inconclusive here, but since the behavior at infinity is a decaying exponential, we can say that the series converges.
