Currently learning about limits and I'd like to check out if I calculated that all correctly:
number 1: SOLVED
$$a_n=\frac{n(n+1)}{(n^2+1)}$$ This was (hopefully) very simple, sadly I stil lack the correct expressions to >!explain it. So I divide by $n_2$: EDITED Nr. 1 AFTER POSTS OF OTHER USERS So sequence a converges to 1: $$\lim_{n \to \infty }a_n=\frac{n(n+1)}{(n^2+1)} =\frac{1}{1}=1$$
number 2: UNSOLVED $$a_n=\frac{c_2n^2+c_1n+c_0}{b_2n^2+b_1n+b_0} \text{with $c_1,c_2,c_3 \in \Bbb{R} \land b_1,b_2,b_3 \in \Bbb{R}_{>0}$}$$ Intuitively I would say this goes towards infinity, so I don't believe it converges. Yet, I find myself having difficulties calculating that.
$$\lim_{n \to \infty }a_n=\frac{c_2n^2+c_1n+c_0}{b_2n^2+b_1n+b_0}$$
Divide by $n^2$:
$$\dfrac{c_2+\dfrac{c_1}{n}+\dfrac{c_0}{n^2}}{b_2+\dfrac{b_1}{n}+\dfrac{b_0}{n^2}}=\dfrac{c_2}{b_2}$$
So even though $b_0$ is positive (by definition), we can not say that the sequence converges for any specific number..? (If I put in $\infty$ for $c_2$ into the sequence it returns $\infty$, if I put it into $c_3$ it returns $-\infty$, if I put it into both result is undefined? or is it infinity again?)
number 3: SOLVED
$$a_n= \frac{13n^6-27}{26n^5\sqrt{n}+39\sqrt{n}}$$ So here I can divide by $n^6$ (I will skip some parts to not clutter up my question any more, everything $\dfrac{1}{n^x}$ and $\sqrt(n)$ turns to 0: $$\lim_{n \to \infty }a_n= \frac{13-0}{\dfrac{26n^{5+0,5}}{n^6}+0} = \frac{13}{\dfrac{26}{n^{0,5}}} = \frac{13}{0} = \infty $$ So again it doesn't have a limit..? easier solution from: https://math.stackexchange.com/a/2791582/560190
For number 1 we have
$$a_n=\frac{n(n+1)}{(n^2+1)}=\frac{(1+1/n)}{(1+1/n^2)} $$
and for number 3
$$a_n= \frac{13n^6-27}{26n^5\sqrt{n}+39\sqrt{n}}=\frac{n^6}{n^5\sqrt{n}}\cdot\frac{13-27/n^6}{26+39/n^5} $$