Determine convergence and limit I want to determine whether each of the sequences converges, and if so, to what limit.
$$\begin{align} 1. \quad \quad & a_n=\frac{n^{2}}{n+1}-\frac{n^{2}+1}{n}\\
2. \quad \quad & a_n=\sqrt{n+1}-\sqrt{n}
\end{align}$$
For the first one, I tried to use ratio test, but it's not working. For the second, I tried to multiply it by $\sqrt{n+1}+\sqrt{n}$ and then devided it. But I'm stuck again and don't know which test should I use to determine its convergence.

For the first sequence, note that
$$\begin{align}
a_n & = \frac{n^2}{n+1} - \frac{n^2+1}{n}\\
& = \frac{n^3-n^3-n-n^2-1}{n^2+n}\\
& = \frac{-n^2-n-1}{n^2+n}.
\end{align}$$
And taking the limit you obtain
$$\lim_{n \to \infty} a_n = -1 \quad (\in \mathbb{R})$$
so it converges.
For the second sequence, as you said, multiply the by $\frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}}$ and you will obtain
$$\begin{align}
a_n & = (\sqrt{n+1}-\sqrt{n}) \cdot \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}}\\
& = \frac{(\sqrt{n+1})^2-(\sqrt{n})^2}{\sqrt{n+1} + \sqrt{n}}\\
& = \frac{n+1-n}{\sqrt{n+1}+\sqrt{n}}\\
& = \frac{1}{\sqrt{n+1}+\sqrt{n}}.
\end{align}$$
Taking the limit you get
$$\lim_{n \to \infty} a_n = 0 \quad (\in \mathbb{R})$$
so it converges too.
