The sequence $a_{n}$ (where $n$ is a natural number) is defined as follows:

$$
a_{0}=2 \quad \text { and } \quad a_{n}=a_{n-1}-\frac{n}{(n+1)!}, \quad \text { if } \quad n>0
$$

Express $a_{n}$ as a function of $n$.
I. solution. Let's calculate the first few terms of the sequence:

$$
a_{1}=2-\frac{1}{2!}=\frac{3}{2}, \quad a_{2}=\frac{3}{2}-\frac{2}{3!}=\frac{7}{6}, \quad a_{3}=\frac{7}{6}-\frac{3}{4!}=\frac{25}{24}
$$

From this, we conjecture that $a_{n}=\frac{(n+1)!+1}{(n+1)!}$. We will prove this by induction.

The statement is true for the first three terms of the sequence. Assume it is true for $n=k$: $a_{k}=\frac{(k+1)!+1}{(k+1)!}$. Using the recursive formula, we get:

$$
\begin{aligned}
a_{k+1} & =a_{k}-\frac{k+1}{(k+2)!}=\frac{(k+1)!+1}{(k+1)!}-\frac{k+1}{(k+2)!}= \\
& =\frac{(k+2)!+(k+2)-(k+1)}{(k+2)!}=\frac{(k+2)!+1}{(k+2)!}
\end{aligned}
$$

This proves that the conjecture is true for all $n$. Therefore, the $n$-th term of the sequence is

$$
a_{n}=\frac{(n+1)!+1}{(n+1)!}=1+\frac{1}{(n+1)!}
$$

II. solution. Notice that

$$
\frac{n}{(n+1)!}=\frac{(n+1)-1}{(n+1)!}=\frac{n+1}{(n+1)!}-\frac{1}{(n+1)!}=\frac{1}{n!}-\frac{1}{(n+1)!}
$$

Consider the differences between consecutive terms of the sequence:

$$
a_{0}-a_{1}=\frac{1}{2!}, \quad a_{1}-a_{2}=\frac{2}{3!}, \quad \ldots, \quad a_{n-1}-a_{n}=\frac{n}{(n+1)!}
$$

Using (1), we can rewrite these differences as follows:

$$
a_{0}-a_{1}=\frac{1}{1!}-\frac{1}{2!}, \quad a_{1}-a_{2}=\frac{1}{2!}-\frac{1}{3!}, \quad \ldots, \quad a_{n-1}-a_{n}=\frac{1}{n!}-\frac{1}{(n+1)!}
$$

Add up these differences, then rearrange the resulting equation (using the fact that $a_{0}=2$):

$$
\begin{array}{r}
a_{0}-a_{n}=1-\frac{1}{(n+1)!} \\
a_{n}=1+\frac{1}{(n+1)!}=\frac{(n+1)!+1}{(n+1)!}
\end{array}
$$